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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 50201. |
A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25^(@)C produces 6.11 litres of CO_(2). Find out the amount of heat evolvedon burning one litre of the mixture. The heatsof combustion of ethylene and methane are - 1423 and - 891kJ mol^(-1) at 25^(@)C |
| Answer» <html><body><p></p>Solution :Combustion <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> of <a href="https://interviewquestions.tuteehub.com/tag/ethylene-976061" style="font-weight:bold;" target="_blank" title="Click to know more about ETHYLENE">ETHYLENE</a> and methane are <br/> `C_(2)H_(4) + 3O_(2)rarr 2CO_(2) + 2H_(2)O , Delta H = - 1423kJ` <br/>`CH_(4) + 2O_(2) rarr CO_(2) + 2H_(2)O , Delta H = - 891 kJ`<br/> Suppose volume of `C_(2)H_(4)` in the mixture of `C_(2)H_(4)` in the mixture`=<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>` litres. Then volume of `CH_(4)` in the mixture `= ( 3.67 -x) ` litres<br/> From the abovereaction `:` <br/> 1 litres of `C_(2) H_(4)` gives `CO_(2)= 2 `litres <br/>`:. ` x litre of `C_(2)H_(4)` gives ` CO_(2) = 2 x `litres <br/> 1 litre of `CH_(4)` gives `CO_(2) = 1 ` litres <br/> `:.( 3.67 - x)` litres of `CH_(4)` gives `CO_(2) = ( 3.67 - x)` litres <br/> Total `CO_(2)` produced `= 2 x + ( 3.67 - x ) = ( 3.67 + xx) ` litres <br/> `:. 3.67+ x = 6.11` or ` x= 2.44` <br/> `:. ` 1 litre of the mixture will contain `C_(2)H_(4) = ( 2.44)/( 3.67 )= 0.66 `litre <br/> and ` CH_(4) = 1 - 0.66 = 0.34 ` litre <br/>Volume of 1 mole `C_(2)H_(4)` or `CH_(4)` at `0^(@)C= 22.4L` <br/>Volume of 1 mole `C_(2)H_(4)` or `CH_(4)` at `25^(@)C = ( 22.4 xx <a href="https://interviewquestions.tuteehub.com/tag/298-1834947" style="font-weight:bold;" target="_blank" title="Click to know more about 298">298</a> ) /( 273)= 24.45 L` <br/>24.45 litres of `C_(2)H_(4)`at `25^(@)C` give heat `= 1423kJ` <br/> `:. 0.66` litre of `C_(2) H_(4)` will give heat`= ( 1423)/( 24.45) xx 0.66 kJ = 3841kJ` <br/> `24.45` litre of`CH_(4)` give heat`= 891 kJ` <br/> `:. ` 0.34litre of` CH_(4)` will give heat`= (891)/(24.45) xx0.34 kJ = 12.39 kJ` <br/> `:. `Total heatproduced `= 38.41 +12.39 =50.38 kJ`</body></html> | |
| 50202. |
A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25^@C and at 1 atm pressure produce 6.11 lit of carbon dioxide. Find out the amount of heat evolved in kJ, during this combustion. (DeltaH_C(CH_4)=-890 "kJ mol"^(-1)) and (DeltaH_C(C_2H_4)=-1423 "kJ mol"^(-1)) . |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Given : <br/> `DeltaH_C(CH_4)=-<a href="https://interviewquestions.tuteehub.com/tag/890-1929934" style="font-weight:bold;" target="_blank" title="Click to know more about 890">890</a> "kJ mol"^(-1)` <br/> `DeltaH_C(C_2H_4)=-1423 "kJ mol"^(-1)` <br/> Let the mixture containx lit of `CH_4` and (3.67-<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>) lit of ethylene . <br/> `underset" x lit"(CH_4)+2O_2 to underset"x lit"(CO_2) + 2H_2O` <br/> `underset"(3.67-x)lit"(C_2H_4 + 3O_2) to underset"2(3.67-x)lit"(2CO_2 + 2H_2O)` <br/> Volume of <a href="https://interviewquestions.tuteehub.com/tag/carbon-16249" style="font-weight:bold;" target="_blank" title="Click to know more about CARBON">CARBON</a> dioxide formed =x +2 (3.67 -x) =6.11 lit <br/> x+7.34 -2x =6.11 <br/> x=1.23 lit <br/> Given mixture <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> 1.23 lit of methane and 2.44 lit of ethylene , hence <br/> `DeltaH_C=[(DeltaH_C(CH_4))/"22.4 lit"xx(x)"lit"]+[(DeltaH_C(C_2H_4))/"22.4 lit"xx(3.67-x)"lit"]` <br/> `DeltaH_C=[(-890"kJ mol"^(-1))/"22.4 lit"xx"1.23 lit"]+[(-1423)/"22.4 lit"xx(3.67-1.23)"lit"]` <br/> `DeltaH_C=[-48.87 "kJ mol"^(-1) ]+[-155 kJ "mol"^(1)]` <br/> `DeltaH_C=-203.87 "kJ mol"^(-1)`</body></html> | |
| 50203. |
A gas mixture contains equal number of molecules of N_(2) and SF_(6) Some of it is passed through gaseous effussion apparatus Calculate how many molcules of N_(2) are pressent in the product gas for every 100 molecules of SF_(6) ? (At wt of F=20) . |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/233-1826012" style="font-weight:bold;" target="_blank" title="Click to know more about 233">233</a></body></html> | |
| 50204. |
A gas mixture contains accetylene and carbondioxide. 20 litres of this mixture required 20 litres of oxygen for complete combustion. If all gases are measured under similar conditions of temperature and pressure, the percentage of acetylene in the mixture is |
| Answer» <html><body><p>`50%`<br/>`40%`<br/>`60%`<br/>`75%`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`C_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)H_(2) = (<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)/(50) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 100` <br/> <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> lit `-(5)/(2)` lit <br/> 20 lit - ? 50 lit <br/> `%C_(2)H_(2) = (20)/(50) xx 100`</body></html> | |
| 50205. |
A gas mixture consisting of 1mol of N_(2) and 3 mol of O_(2) has a pressure of 2 atmm at 0^(@)C. Keeping the volume and the temperature of the mixture constant, some amount of O_(2) was removed from the mixture. As a result, the total pressure of the mixture and the partial pressure of N_(2) in the mixture became 1.5 atm and 0.5 atm respectively. the amount of oxygen gas removed was- |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> g<br/>16 g<br/>32 g<br/>64 g</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 50206. |
A gas mixture at 27^(@)C and 1 atm contains equal masses of He,H_(2),CO_(2) and CH_(4). How do their molecular velocities compare? |
| Answer» <html><body><p>`He=H_(2)=CO_(2)=CH^(4)`<br/>`HeltH_(2)ltCO_(2)ltCH_(4)`<br/>`H_(2)HeltCH_(4)ltCO_(2)`<br/>`CO_(2)ltCH_(4)ltHeltH_(2)`</p>Answer :d</body></html> | |
| 50207. |
A gas is initially at 1 atm pressure. To compress it to 1/4 th of its initial volume, pressure to be applied is:1 atm, 2 atm, 3 atm, 1/4 atm |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a><br/>2 atm<br/><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> atm<br/>1/4 atm.</p>Solution :4 atm</body></html> | |
| 50208. |
A gas is filled into a bulb connected to anopen limb manometer. The level of mercury in the open arm is 2.1 cm lower than that in the other arm of the manoneter. The atmospheric pressure is 740 mm. What is the pressure of the gas ? |
| Answer» <html><body><p></p>Solution :As level in the open arm of the manometer is lower that that in the arm connected to the gas <a href="https://interviewquestions.tuteehub.com/tag/bulb-905352" style="font-weight:bold;" target="_blank" title="Click to know more about BULB">BULB</a>, this means that atmospheric <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> is greater than pressure of the gas by difference of the two <a href="https://interviewquestions.tuteehub.com/tag/levels-1072811" style="font-weight:bold;" target="_blank" title="Click to know more about LEVELS">LEVELS</a>. <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, pressure of gas=740-21=719 <a href="https://interviewquestions.tuteehub.com/tag/mm-1098795" style="font-weight:bold;" target="_blank" title="Click to know more about MM">MM</a>.</body></html> | |
| 50209. |
A gas is filled in a container whose mass is 916.4 g. The mass of the gas plus container is found to be 917.64g. If the container can hold 1107 cm^3 of water, find the density of the gas. |
| Answer» <html><body><p><br/></p>Answer :`<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.1 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) g <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>^(-1)`</body></html> | |
| 50210. |
a gas is enclosed in a room. The pressure, density, temperature and number of moles are p atm, g cm^(-3), t^(@)C andn moles repectively. What will be the pressure, temperature, density and number of moles, in each compartment. If room is partitioned into four equal compartments ? |
| Answer» <html><body><p></p>Solution :Pressure in each compatment is same. P. <br/> <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> will remain same `t^(@)C`. <br/> Density will remain same. `d <a href="https://interviewquestions.tuteehub.com/tag/gcm-467949" style="font-weight:bold;" target="_blank" title="Click to know more about GCM">GCM</a>^(-3)`<br/> The number of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> in each <a href="https://interviewquestions.tuteehub.com/tag/component-926634" style="font-weight:bold;" target="_blank" title="Click to know more about COMPONENT">COMPONENT</a> will be, `(<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>)/(4)` moles.</body></html> | |
| 50211. |
A gas is enclosed in a room. The temperature, pressure, density and the number of moles respectively are t^@C, p " atm, d g " cm^(-3)and n moles. What will be the pressure, temperature, density and number of moles in each compartment if the room is partitioned into four equal compartments ? |
| Answer» <html><body><p></p>Solution :When the wall between compartments 1 and <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> is removed : <br/> Pressure = p atm. This is because the number of moles and volume of gas will <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> to the same <a href="https://interviewquestions.tuteehub.com/tag/extent-981159" style="font-weight:bold;" target="_blank" title="Click to know more about EXTENT">EXTENT</a>, i.e., will get doubled. <br/> <a href="https://interviewquestions.tuteehub.com/tag/density-17451" style="font-weight:bold;" target="_blank" title="Click to know more about DENSITY">DENSITY</a> = dg `<a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>^(-3)` . Density of the gas remains unchanged because both mass and volume get doubled. <br/> Number of moles = `n/4+n/4=n/2`.</body></html> | |
| 50212. |
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy DeltaU of the gas in joules will be |
| Answer» <html><body><p>`-500 J`<br/>`-505 J`<br/>`+505 J`<br/>`1136.25 J`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/deltau-2053495" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAU">DELTAU</a> = q + <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>` <br/> For adiabatic process q = <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a><br/> `DeltaU = -w = -pDeltaV` <br/> ` = -2.5 atm (4.5 - 2.5)L` <br/> ` =-5 <a href="https://interviewquestions.tuteehub.com/tag/latm-1069541" style="font-weight:bold;" target="_blank" title="Click to know more about LATM">LATM</a> = -5 xx 101.3 J = -506.5J`</body></html> | |
| 50213. |
A gas is allowe to expandin a well insulated container againsta constant external pressure of2.5 atmfrom an initial volume of2.50 Lto a final volume of 4.50L. The change in internal energy DeltaU of thegas in Joules will be |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>1136.25J`<br/>`-<a href="https://interviewquestions.tuteehub.com/tag/500j-1893997" style="font-weight:bold;" target="_blank" title="Click to know more about 500J">500J</a>`<br/>`-505J`<br/>`+505J`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`DeltaU =q+w` <br/> For well insulated <a href="https://interviewquestions.tuteehub.com/tag/container-20566" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINER">CONTAINER</a>, process is adiabatic , `q=0` <br/> `:. DeltaU =w= - P DeltaV`<br/> `= - 2.5 atm ( 4.5 - 2.5) L`<br/> `=- 5L atm` <br/> `= 5xx 101.3 J= - 506.5 J ` <br/> `= -505J`</body></html> | |
| 50214. |
A gas heated to 0^(@)C to 546^(@)C at 5 bar pressure at 5 bar pressure it its volume becomes the third then calculate final pressure. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/45-316951" style="font-weight:bold;" target="_blank" title="Click to know more about 45">45</a> <a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a></body></html> | |
| 50215. |
A gas has molecular formula (CH)_n.If vapour density of the gas is 39, what should be the formulaof the compound ? |
| Answer» <html><body><p>`C_3H_3`<br/>`C_4H_4`<br/>`C_2H_2`<br/>`C_6H_6`<br/></p>Solution :Mol. Wt. `=2xxV.D. = 2xx39=78` <br/> `(CH)_n=(<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>)_n " or " 13xxn=78` <br/> <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>=78/13=6 <br/> `therefore " <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> formula " = C_6H_6`</body></html> | |
| 50216. |
A gas has density 1.52gL^(-1)at 1.5atm. Calculate its RMS velocity. |
| Answer» <html><body><p></p>Solution :`1.732 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) <a href="https://interviewquestions.tuteehub.com/tag/cms-920094" style="font-weight:bold;" target="_blank" title="Click to know more about CMS">CMS</a>^(-1)`</body></html> | |
| 50217. |
A gas fired power plant burns methane, CH_(4)(g) for which standard enthalpy of combustion is -890 kJ mol^(-1). How many moles of CO_(2) gas are produced for every 1.0MJ ( mega joule) of energy produced by it ? |
| Answer» <html><body><p><br/></p>Solution :`1 <a href="https://interviewquestions.tuteehub.com/tag/mj-548176" style="font-weight:bold;" target="_blank" title="Click to know more about MJ">MJ</a> = <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>) <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>`</body></html> | |
| 50218. |
A gas formed by the action of alcohlic KOH on ethyl iodide, decolourless alkaline KMnO_(4) solution. The gas is |
| Answer» <html><body><p>`CH_(4)`<br/>`C_(2)H_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)`<br/>`C_(2)H_(4)`<br/>`C_(2)H_(2)`</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 50219. |
A gas filled in a bulb of capacity 25.1 mL at 27°C and 750 mm pressure weighs 0.072 g. If 1 litre of hydrogen at S.T.R weighs 0.09 g, calculate the molecular mass of the gas. |
| Answer» <html><body><p></p>Solution :Let the volume of the gas at STP be `V_(2)` mL <br/> `P_(1) = 750 <a href="https://interviewquestions.tuteehub.com/tag/mm-1098795" style="font-weight:bold;" target="_blank" title="Click to know more about MM">MM</a>, V_(1) = 25.1 mL, T_(1) = <a href="https://interviewquestions.tuteehub.com/tag/27-298706" style="font-weight:bold;" target="_blank" title="Click to know more about 27">27</a>^(@) <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 300 K`<br/> `P_(2) = 760 mm, V_(2) = ?, T_(2) = 273 K` <br/> According to the gas equation: <br/> `(P_(1)V_(1))/T_(1) = (P_(2)V_(2))/T_(2)` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2))` <br/> Substituting the values, we get <br/> `V_(2) = (750 xx 25.1 xx 273)/(300 xx 760)` <br/> = 22.5 mL <br/> Mass of 22.5 mL of `H_(2)` at S.T.P. <br/> `=0.09/1000 xx 22.5 = 2.025 xx 10^(-3) <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>` <br/> V.D. = `("Mass of 22.5 mL of gas at S.T.P")/("Mass of 22.5 mL of H_(2) at S.T.P")` <br/> `=0.072/(2.025 xx 10^(-3))` <br/> = 35.55 <br/> Hence, <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> mass `=2 xx V.D. = 2 xx 35.55 = 71.1`</body></html> | |
| 50220. |
A gas filled freely callapsible ballon is pushed from the surface level of take to a depth of 100 meter Calculate what per cent of its original volume the balloon finally have ? Assume ideal gas nature . |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`9.37%ofV_(i)`</body></html> | |
| 50221. |
A gas expands isothermally and reversibly. The work done by the gas is |
| Answer» <html><body><p>Zero<br/>Maximum <br/>Minimum<br/>Not known</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 50222. |
A gas expands from a volume of 1m^(3) to a volume of2m^(3) against an external pressure of10^(5) Nm^(-2). The work done by the gas will be |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(5) kJ `<br/>` 10^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) kJ`<br/>` 10^(2) <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>`<br/>`10^(3)J`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`w = - PDeltaV =- 10^(5) <a href="https://interviewquestions.tuteehub.com/tag/nm-579234" style="font-weight:bold;" target="_blank" title="Click to know more about NM">NM</a>^(-2) ( 2-1) m^(3) =- 10^(5) Nm = - 10^(5) J = -10^(2) kJ `</body></html> | |
| 50223. |
A gas expands against a variable pressure given by P = (20)/(V) (where P in atm and V in L). During expansion from volume of 1 litre to 10 litre, the gas undergoes a change in internal energy of 400 J. How much heat is absorbed by the gas during expansion? |
| Answer» <html><body><p>46 J<br/>4660 J<br/>5065.8 J<br/>4260 J</p>Solution :`intdw=int-P.<a href="https://interviewquestions.tuteehub.com/tag/dv-433533" style="font-weight:bold;" target="_blank" title="Click to know more about DV">DV</a>`<br/> `"<a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a>"w=-int20.(dV)/(V)=-20ln.(V_(2))/(V_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))`<br/> `w=-46.06 " L-atm"=-4665.8 J`<br/> `DeltaU=q+wimplies400=q-4665.8`<br/>`q = 5065.8 J`</body></html> | |
| 50224. |
A gas expands against a constant external pressure so thatthe work done is 607.8 J. The work done in litre atmosphere is |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a></body></html> | |
| 50225. |
A gas diffuses one-thrid as fast as O_(2) at 100^(@)C. This gas could be: |
| Answer» <html><body><p>He(M=4)<br/>`C_(2)H_(5)(M=48)`<br/>`C_(7)H_(<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>)(M=96)`<br/>`C_(5)F_(12)(M=288)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :d</body></html> | |
| 50226. |
A gas diffuses four times as quickly as oxygen. The molecular weight of the gas is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 50227. |
A gas described by van der Waals equation |
| Answer» <html><body><p>behaves similar to an ideal gas in the limit gas in the limit of large molar volumes<br/>behaves similar to an ideal gas in the limit of large pressures <br/>is characterised by van der Waals coefficients that are dependent on identity of the gas but are independent of the pressure<br/>has the pressure that is lower than the pressure <a href="https://interviewquestions.tuteehub.com/tag/exerted-7272976" style="font-weight:bold;" target="_blank" title="Click to know more about EXERTED">EXERTED</a> by the same behaving ideally </p>Solution :van der Waals equation is <br/> `(P+(an^(2))/(V^(2)))(V-nb)=nRT`.<br/> (a) is correct because for large molar volumes, `an^(2)//V^(2)` is small and can be neglected in comparison to P and nb can be neglected in comparison to V. Eqn. becomes PV=nRT. <br/> (c ) is correct because van der Waals coefficients depend only on the <a href="https://interviewquestions.tuteehub.com/tag/nature-1112013" style="font-weight:bold;" target="_blank" title="Click to know more about NATURE">NATURE</a> of the gas. <br/> (d) is correct because due to inward pull on the <a href="https://interviewquestions.tuteehub.com/tag/molecules-563030" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULES">MOLECULES</a> striking the <a href="https://interviewquestions.tuteehub.com/tag/wall-1448655" style="font-weight:bold;" target="_blank" title="Click to know more about WALL">WALL</a>, the pressure is lower.</body></html> | |
| 50228. |
A gas decolourises alkaline KMnO_4 solution but does not give precipitate with AgNO_3, it is |
| Answer» <html><body><p>`CH_4`<br/>`C_2H_4`<br/>`C_2H_2`<br/>`C_2H_6`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`C_2H_4 (CH_2=CH_2)` <a href="https://interviewquestions.tuteehub.com/tag/decolourises-2567646" style="font-weight:bold;" target="_blank" title="Click to know more about DECOLOURISES">DECOLOURISES</a> `KMnO_4` solution but does not <a href="https://interviewquestions.tuteehub.com/tag/give-468520" style="font-weight:bold;" target="_blank" title="Click to know more about GIVE">GIVE</a> <a href="https://interviewquestions.tuteehub.com/tag/ppt-592533" style="font-weight:bold;" target="_blank" title="Click to know more about PPT">PPT</a>. with `AgNO_3` solution.</body></html> | |
| 50229. |
A gas cylinder contains oxygen gas at 25^@C and 15.0 atm. If the temperature of the surroundings rises to 42^@C, whatwould be the pressure of the gas in the cylinder ? |
| Answer» <html><body><p></p>Solution :In the present <a href="https://interviewquestions.tuteehub.com/tag/case-910082" style="font-weight:bold;" target="_blank" title="Click to know more about CASE">CASE</a>, <br/> `P_1 = 15.0 " atm,"T_1 = 25^@<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 25 + <a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a> = 298 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>, P_2 = ?, T_2 = <a href="https://interviewquestions.tuteehub.com/tag/42-316129" style="font-weight:bold;" target="_blank" title="Click to know more about 42">42</a>^@C = 42 + 273 = 315 K` <br/> Using the pressure-temperature law `P_1/T_1 = P_2/T_2`and substituting the values, we get <br/> `P_2 = (P_1T_2)/(T_1) = (15.0 x 315)/(298) = 15.86` atm</body></html> | |
| 50230. |
A gas cylinder containing cooking gas can withstand a pressure of 14.9 atmospheres. The pressure gauge of the cylinder indicates 12 atmosphere at 27^(@)C. Due to sudden fire in the building, the temperature starts rising.At what temperature the cylinder will explode ? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`P_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`=12 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>, `T_(1)`=300 K, `P_(2)`=14.9 atm, `T_(2)`=?<br/> As volume of the <a href="https://interviewquestions.tuteehub.com/tag/cylinder-942617" style="font-weight:bold;" target="_blank" title="Click to know more about CYLINDER">CYLINDER</a> is constant, `(P_(1))/(T_(1))=(P_(2))/(T_(2))`</body></html> | |
| 50231. |
A gas cylinder can withstand a pressure of 15 bar. The pressure gauze indicates 12 bar at 27^(@)C. IF the building catches fire suddenly. At what temperature will the cylinder explode ? |
| Answer» <html><body><p></p>Solution :`P_(1) = 12 " bar"P_(2) =`bar <br/> `T_(1) = 273 + <a href="https://interviewquestions.tuteehub.com/tag/27-298706" style="font-weight:bold;" target="_blank" title="Click to know more about 27">27</a> = 300 K ""T_(2) = ?` <br/> `(P_(1))/(T_(1)) = (P_(2))/(T_(2)) ""T_(2) = (P_(2) xx T_(1))/(P_(1)) = (15 xx 300)/(12)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> at which cylinder explodes = <a href="https://interviewquestions.tuteehub.com/tag/375-309985" style="font-weight:bold;" target="_blank" title="Click to know more about 375">375</a> K = `102^(@)C`</body></html> | |
| 50232. |
A gas cylinder withstands a pressure of 14.9 atm. Its pressure gauze indicates 12 atm at 27^@C. If the building catches fire suddenly, at what temperature the cylinder explodes? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :above `99.5^@<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>`</body></html> | |
| 50233. |
A gas cylinder can withstand a pressure of 14.9 bar at 27^@C. The pressure gauze indicates 12 bar. If the building catches fire suddenly, at what temperature the cylinder expodes? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :above `99.5^@ `<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 50234. |
A gas mixture contains acetylene and carbondioxide. 20 lit of this mixture requires 20 lit of oxygen under the same conditions for complete combustion. The percentage by volume of acetylene in the mixture is |
| Answer» <html><body><p>0.5<br/>0.4<br/>0.6<br/>0.75</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 50235. |
A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure or 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system. |
| Answer» <html><body><p></p>Solution :Given <a href="https://interviewquestions.tuteehub.com/tag/data-25577" style="font-weight:bold;" target="_blank" title="Click to know more about DATA">DATA</a> q = 400 J, `V_(1) = 5L , V_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) = <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> L`<br/>`Delta <a href="https://interviewquestions.tuteehub.com/tag/u-1435036" style="font-weight:bold;" target="_blank" title="Click to know more about U">U</a>` = q - <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a> (heat is given to the system (+q), work is done by the system (-w)<br/>`Delta u = q - Pd V `<br/>= 400 J - 1 atm ( 10 - 5) L<br/>= 400 J - 5 atm L[`:'`1 L atm = 101 . 33 J ]<br/>= 400 J - ` 5 xx 101 . 33 ` J<br/>= 400 J - 506 65 J<br/>= - 106 . 65 J .</body></html> | |
| 50236. |
A gas can expand from 100 ml to 250 ml under a constant pressure of 2 atm. The work done by the gas is |
| Answer» <html><body><p>30.38 <a href="https://interviewquestions.tuteehub.com/tag/joule-526299" style="font-weight:bold;" target="_blank" title="Click to know more about JOULE">JOULE</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> Joule <br/><a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> k/Joule<br/>16 Joule </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 50237. |
A gas can expand from 100 ml to 250 ml under a constant pressure of 2 atm. The work done by the gas is _____ |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/deltav-2053504" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAV">DELTAV</a>`=expansion in <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> =100 to 250 <br/> `DeltaV=V_2-V_1`=250-100 <br/> `DeltaV`=150 ml =0.15litre <br/> Work done = ? <br/> Pressure =2 atm <br/> `w=-P DeltaV` <br/> `=-2xx0.15 "<a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a>" xx 101.3 J <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(-1) "atm"^(-1)` <br/> =-30.39 J</body></html> | |
| 50238. |
A gas can be liquefied if : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/forces-16875" style="font-weight:bold;" target="_blank" title="Click to know more about FORCES">FORCES</a> of attraction are <a href="https://interviewquestions.tuteehub.com/tag/low-537644" style="font-weight:bold;" target="_blank" title="Click to know more about LOW">LOW</a> under <a href="https://interviewquestions.tuteehub.com/tag/ordinary-1138437" style="font-weight:bold;" target="_blank" title="Click to know more about ORDINARY">ORDINARY</a> conditions <br/>forces of attraction are high under ordinary conditions <br/>forces of attraction are zero under ordinary conditions <br/>forces of attraction either high or low under ordinary conditions </p>Solution :As long <a href="https://interviewquestions.tuteehub.com/tag/imf-498820" style="font-weight:bold;" target="_blank" title="Click to know more about IMF">IMF</a> is <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a>, gas can be liquified.</body></html> | |
| 50239. |
A gas can be liquefied by |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/cooling-933550" style="font-weight:bold;" target="_blank" title="Click to know more about COOLING">COOLING</a> <br/>Compressing <br/>Both <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> and <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> <br/>None of these </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/applying-1982651" style="font-weight:bold;" target="_blank" title="Click to know more about APPLYING">APPLYING</a> P below `T_C`.</body></html> | |
| 50240. |
A gas can be liquefied by pressure alone when its temperature is |
| Answer» <html><body><p>Higher than its critical temperature <br/><a href="https://interviewquestions.tuteehub.com/tag/lower-1080637" style="font-weight:bold;" target="_blank" title="Click to know more about LOWER">LOWER</a> than its critical temperature <br/><a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> to its critical temperature <br/>Equal to its Boyle's temperature </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/liquified-2158355" style="font-weight:bold;" target="_blank" title="Click to know more about LIQUIFIED">LIQUIFIED</a> only below `T_C`.</body></html> | |
| 50241. |
A gas bulb containing air is connected to an open limb manometer at 27^(@) C and at 750 mm Hg . Assuming that intially the level of Hg in the both limbs were same . The bulb was heated to 77^(@)C , what will be differences in the levels of Hg in two limbs ? (Assuming the volume difference of the gas produced is negligible at higher temperature) . |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>7.5 cm Hg <br/>8 cm Hg <br/>6 cm Hg <br/>12.5 cm Hg </p>Solution :`because (P_(1))/(T_(1)) = (P_(2))/(T_(2)) , (<a href="https://interviewquestions.tuteehub.com/tag/750-335306" style="font-weight:bold;" target="_blank" title="Click to know more about 750">750</a>)/(300) = (P_(2))/(350)` <br/> `therefore P_2 = 875` mm Hg <br/> `therefore` So , difference in height= `875 - 750 = <a href="https://interviewquestions.tuteehub.com/tag/125-271180" style="font-weight:bold;" target="_blank" title="Click to know more about 125">125</a>` mm <br/> Hg = 12.5 cm of Hg</body></html> | |
| 50242. |
A gas bulb of 1 litre capacity contains 2.0 xx 10^(21) molecules of nitrogen exerting a pressure of 7.57 xx 10^3 Nm^(-2) . Calculate the root mean square speed and the temperature of gas molecules. If the ratio of most probable speed to the root mean square speed is 0.82, calculate the most probable speed for the molecules at this temperature. |
| Answer» <html><body><p><br/></p>Solution :Number of moles of <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a> = `("number of molecules")/(6.023 xx 10^(23))` <br/> `= (2.0 xx 10^(11))/(6.023 xx 10^(23)) = 3.32 xx 10^(-3)` <br/> According to the gas equation PV = <a href="https://interviewquestions.tuteehub.com/tag/nrt-1109890" style="font-weight:bold;" target="_blank" title="Click to know more about NRT">NRT</a><br/> In the <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> case, `P = 7.57 xx 10^3 N m^(-2), <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> = 1` <br/> `L = 10^(-3) m^(3)`, <br/> `R = 8.314 J K^(-1) mol^(-1)` T = ?<br/> Substituting the values, we have <br/> `7.57 xx 10^3 xx 10^(-3) = 3.32 xx 10^(-3) xx 8.314 <a href="https://interviewquestions.tuteehub.com/tag/xxt-3292708" style="font-weight:bold;" target="_blank" title="Click to know more about XXT">XXT</a>` <br/> `:. "" T= 274.2 K` <br/> RMS speed is given by <br/>`u = sqrt((3RT)/M)= sqrt((3 xx 8.314 xx 274.2)/(28 xx 10^(-3)))= 494.2 m s^(-1)` <br/> (Molar mass of `N_2 " in kg " = 28 xx 10^(-3)`) <br/> `:. " Most probable speed " = 0.82 xx RMS` speed <br/> `:. " Most probable speed of " N_2` under given conditions <br/> `= 0.82 xx 494.2 = 405.2 m s^(-1)`</body></html> | |
| 50243. |
A gas believed to be the cause of explosin in coal mine is |
| Answer» <html><body><p>`CH_4`<br/>`C_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)H_6`<br/>`C_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)H_(8)`<br/>`C_(4)H_(10)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 50244. |
A gas at a pressure of 5 atm is heated from 0^(@)C" to "546^(@)C and is simultaneously compressed to one third of its original volume. Find the final pressure of the gas. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/45-316951" style="font-weight:bold;" target="_blank" title="Click to know more about 45">45</a> <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>`</body></html> | |
| 50245. |
A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same conditions. The correct option above the gas and its compressibility factor (Z) is : |
| Answer» <html><body><p>`Z lt1` and attractive forces are dominant<br/>`Z lt 1` and <a href="https://interviewquestions.tuteehub.com/tag/repulsive-613883" style="font-weight:bold;" target="_blank" title="Click to know more about REPULSIVE">REPULSIVE</a> forces are dominant<br/>`<a href="https://interviewquestions.tuteehub.com/tag/zgt1-2345918" style="font-weight:bold;" target="_blank" title="Click to know more about ZGT1">ZGT1</a>` and attractive forces are dominant<br/>`<a href="https://interviewquestions.tuteehub.com/tag/zgt-3893032" style="font-weight:bold;" target="_blank" title="Click to know more about ZGT">ZGT</a> 1` and repulsive forces are dominant.</p>Solution :`Z lt1` and attractive forces are dominant</body></html> | |
| 50246. |
A gas at 0^@C and 1 atmospheric pressure occupies 2.5 litres. What change in temperature would be necessary if the pressure is to be adjusted to 1.5 atmospheres and the gas has been transferred to a 2.0 litre container? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`54.6^@<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>`</body></html> | |
| 50247. |
A gasabsorbs photonof 355nm andemits two wavelengthif oneof theemissionis at 680nm the otheris at . |
| Answer» <html><body><p>1035 <a href="https://interviewquestions.tuteehub.com/tag/nm-579234" style="font-weight:bold;" target="_blank" title="Click to know more about NM">NM</a> <br/> 325 nm <br/> 743 nm <br/> 518 nm</p>Solution :`E= hv = (hc)/(lambda)` <br/> `(hc)/(lambda) = (hc )/(lambda)+ (hc)/( lambda_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))` <br/> `(1)/(<a href="https://interviewquestions.tuteehub.com/tag/355-309048" style="font-weight:bold;" target="_blank" title="Click to know more about 355">355</a>)=(1)/(680)+ (1)/(lambda)` <br/> `=(680)/(355) xx (355)/( 680)` <br/> `lambda_(2)= (355 xx 680)/( 325)= 742.769 = 743nm`</body></html> | |
| 50248. |
A gas absorbs a photon of 355 nm and emit at two wavelengths. If one of the emission is at 680 |
| Answer» <html><body><p>325 nm<br/>743 nm<br/>518 nm<br/>1035 nm</p>Solution :As energies are additive, `E = E_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) + E_(2)` <br/> or `(<a href="https://interviewquestions.tuteehub.com/tag/hc-1016346" style="font-weight:bold;" target="_blank" title="Click to know more about HC">HC</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/lamda-536483" style="font-weight:bold;" target="_blank" title="Click to know more about LAMDA">LAMDA</a>) = (hc)/(lamda_(1)) + (hc)/(lamda_(2)) or (1)/(lamda) = (1)/(lamda_(1)) + (1)/(lamda_(2))` <br/> But `lamda = 355 nm, lamda_(1) = <a href="https://interviewquestions.tuteehub.com/tag/680nm-1911337" style="font-weight:bold;" target="_blank" title="Click to know more about 680NM">680NM</a>,` <br/> `:. (1)/(355) = (1)/(680) + (1)/(lamda_(2))` <br/> or `(1)/(lamda_(2)) = (1)/(355) - (1)/(680) = (325)/(355 xx 680)` <br/> or `lamda_(2) = (355 xx 680)/(325) = 745nm`</body></html> | |
| 50249. |
A gas absorbs 400 J of heat and expands by 2 xx 10^(-3)m^(3) against a constant pressure of 1 xx 10^5 Nm^(-2). The change in internal energy is (1 L atm = 100 J) |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a> <br/>`<a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a>`<br/>`-<a href="https://interviewquestions.tuteehub.com/tag/600j-329888" style="font-weight:bold;" target="_blank" title="Click to know more about 600J">600J</a>`<br/>`-200J`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :B</body></html> | |
| 50250. |
A gas absorbs 100 calories of heat energy and is compressed from 10L to 5L by applying an external pressure of 2 atm. Change in internal energy in calories will be nearly. |
| Answer» <html><body><p>312<br/>342<br/>426<br/>562</p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> <a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a> = <a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a> + <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>`</body></html> | |