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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
A pre-weighes vessel was filled with oxygen at N.T.P. and weighed. It was then evacuated, filled with `SO_(2)` at the same temperature and pressure, and again weighted. The weight of oxygen will beA. The same as that of `SO_(20`B. `(1)/(2)` that of `SO_(2)`C. Twice that of `SO_(2)`D. One fourth that of `SO_(2)` |
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Answer» Correct Answer - B `("M. wt. of " O_(2))/("M.wt . of "SO_(2))implies (M_(1))/(M_(2))implies(32)/(64)=(1)/(2)` The weight of oxygen will be `(1)/(2)` that of `SO_(2)` |
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| 102. |
What will be the pressure exerted by a mixture of `3.2 g` of methane and `4.4 g` of methane and `4.4g` of carbon dixide contained in a `9 dm^(3)` flask at `27^(@)C` ? . |
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Answer» `PV =nRT` `n =n_(CH_4) + n_(CO_2) = (3.2)/(16) + (4.4)/(44) =0.3` `V =9dm^(3) =9litre` `:. P xx 9 = 0.3 xx 0.082 xx 300` `:. P =0.82 atm = 0.82 xx 1.013xx 10^(5) Pa` `= 8.31 xx 10^(4) Pa` . |
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| 103. |
The vander waals constant for gases `A,B` and `C` are as follows: `{:(Gas,a//dm^(6)KPa mol^(-2),b//dm^(3) mol^(-1)),(A,405.3,0.027),(B,1215.9,0.030),(C,607.95,0.032):}` Which gas has (i) Highest critical temperature (ii) The largest molecular volume (iii) Most ideal behaviour around `STP`? |
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Answer» `T_(c) = (8a)/(27Rb)` Since, `R` is constant, higher the value of `a//b`, higher will be critical temperature, `V_(c) = 3b` and `V_(c) prop V_(m)` (for a particular gas) therefore higher the value of `V_(c)`, higher will be molar volume of the gas. If the critical temperature is close to `273K`, gas will behave ideally around the `STP`. Let us illustrate teh result in a tabular form. `{:(Gas,a//dm^(6)KPa mol^(-2),b//dm^(3)mol^(-1),T_(c),V_(c),a//b),(A,405.3,0.027,534.97K,0.081,1.501xx10^(4)),(B,1215.9,0.030,1444.42K,0.09,4.053xx10^(4)),(C,607.95,0.032,677.07K,0.096,1.89xx10^(4)):}` (i) `B` gas has the largest critical temperature. (ii) `C` gas has the largest molecular volume. (iii) `A` gas has the most ideal behaviour around `STP` |
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| 104. |
Choose the correct alternative (more than one may be correcT) (B.M.C=Bimolecular collision) (at constant P) (n is constant throughout).A. `lambda` is constant.B. BMC made by 1 molecule per second is directly proportional to T.C. BMC for all the molecuels per unit volume is directly proportional to `T^(2)`D. none of these |
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Answer» Correct Answer - d |
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| 105. |
The most probable kinetic energy of a gas moleculeA. `(kT)/(2)`B. `kT`C. 2kTD. RT |
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Answer» Correct Answer - a |
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| 106. |
The van der Waals parameters for gases W, X, Y and Z are `{:("Gas",a("atm"L^(2)mol^(-2)),b(L"mol"^(-1))),(W," "4.0," "0.027),(X," "8.0," "0.030),(Y," "6.0," "0.032),(Z," "12.0," "0.027):}` Which one of these gases has the highest critical temperature?A. WB. XC. YD. Z |
| Answer» Correct Answer - d | |
| 107. |
The van der Waals parameters for gases W, X, Y and Z are `{:("Gas",a("atm"L^(2)mol^(-2)),b(L"mol"^(-1))),(W," "4.0," "0.027),(X," "8.0," "0.030),(Y," "6.0," "0.032),(Z," "12.0," "0.027):}` Which one of these gases has the highest critical temperature?A. `W`B. `X`C. `Y`D. `Z` |
| Answer» Correct Answer - D | |
| 108. |
van der Waals constant b of helium is 24 mL `mol^(-1)`. Find molecular diameter of helium.A. `1.335xx10^(-10)"cm" `B. `1.335xx10^(-8)"cm" `C. `2.67xx10^(-8)"cm" `D. `4.34xx10^(-8)"cm" ` |
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Answer» Correct Answer - c (c) b=4`xx`volume occupied by molecules of 1 mol of a gas, `or" "b=4N_(A)((4)/(3)pir^(3))` `or" " r=((3b)/(16N_(A)pi))` `=[(3xx24)/(16xx6.023xx10^(23)xx3.14)]^(1//3)" cm" ` `=1.355xx10^(-8) "cm" ` Now molecular diameter, d =2r `=2.67xx10^(-8)"cm"` |
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| 109. |
van der Waals constant b of helium is 24 mL `mol^(-1)`. Find molecular diameter of helium.A. `1.335 xx 10^(-10)cm`B. `1.335 xx 10^(-8)cm`C. `2.67 xx 10^(-8)cm`D. `4.34 xx 10^(-8)cm` |
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Answer» Correct Answer - C `b = 4 xx` volume occuoied by molecules of `1mol` of a gas. or `b = 4N_(A) ((4)/(3)pir^(3))` or `r = ((3b)/(16N_(A^(pi))))^(1//3)` `= [(3xx24)/(16xx6.023 xx 10^(23)xx3.14)]^(1//3)cm` `= 1.355 xx 10^(-8)cm` Now, molecular diameter, `d = 2r = 2.67 xx 10^(-8)cm` |
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| 110. |
One mole of an ideal monoatomic gas is mixed with 1 mole of an ideal diatomic gas The molar specific heat of the mixture at constant volume is .A. `3cal`B. `4cal`C. `8cal`D. `9cal` |
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Answer» `C_(v) = (3)/(2)R` (for monoatomic) and `(5)/(2)R` (for diatomic) Thus for mixture `C_(v)=[[(3)/(2)R+(5)/(2)R]]/(2) =2R =4cal` . |
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| 111. |
A vessel of vossel of volume `8.0 xx 10^(-3) m^(-3)` contains ideal gas at `300 K` and `200kPa` The gas is allowed to leak till the pressure falls to `125 kPa` Calculate the mole of gas leaked out if temperature remains constant Assume ideal nature . |
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Answer» At ` t = 0 n_(g)` ` =(PV)/(RT) = (200 xx 10^(3) xx 8.0 xx 10^(-3))/(8.314 xx 300) = 0.64` After leakage, `n_(g)` left ` = (PV)/(RT) = (125 xx 10^(3) xx 8.0 xx 10^(-3))/(8.314 xx 300) =0.64` `:. ng` leaked out `= 0.64 -0.40 = 0.24` . |
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| 112. |
The molar volume of He at `10 .1325MP` and `273K` is 0.011075 of its molar volume at 101.325 kPa at `273 K` Calculate the radius of helium atom The gas is assumed to show real nature Neglect the value of a for He . |
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Answer» For real gas `[P+(a)/(V^(2))][V -b] =RT` or `P[V -b] =RT` (neglectinga) `:. (10.1325 xx10^(6))/(101325) [V_(1)-b] =0.0821 xx 273` `(101325 Pa = 1atm)` or `100[V_(1)-b] = 0.0821 xx 273 =22.41 ..(1)` `(101.325 xx 10^(3))/(101325) [V_(2)-b] = 0.0821 xx 273` or `[V_(2) -b] =22.41 ...(2)` By eq.`(1) V_(1) = 0.2241 +b ...(3)` By eq (2) `V_(2) = 22.41 +b...(4)` By eqs (3) and `(4) (V_(1))/(V_(2)) = (0.2241 +b)/(22.41 +b)` `(0.011075V_(2))/(V_(2)) = (0.2241+b)/(22.41 +b)` `(V_(1) = 0.011075 V_(2)` is given) `:. b = 0.024 litre mo1^(-1) = 24 cm^(3) mo1^(-1)` `:. b =4N xx upsilon = 4 xx 6.023 xx 10^(23) xx (4)/(3) pi r^(3)` or `24 = 4 xx 6.023 xx 10^(23) xx (4)/(3) xx (22)/(7) xx r^(3)` `:. r = 1.33 xx 10^(-8) cm` . |
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| 113. |
Calculate the average volume available to a molecule in a sample of `N_(2)` at `NTP` Also report the average distance between two neighbouring molecules if a nitrogen molecule is assumed to be spherical Comment on the result if the radius of one `N_(2)` molecule ` =2 xx 10^(-8) cm^(3)` . |
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Answer» Volume occupied by `N` molecules of `N_(2) = 22400 cm^(3)` `:.` Volume occupied by 1 molecules of `N_(2)` ` = (22400)/(6.02 xx 10^(23)) = 3.72 xx 10^(-20) cm^(3)` Also volume of 1 molecule of `N_(2) = (4)/(3) pi r^(3)` `:. (4)/(3) xx (22)/(7) xx r^(3) = 3.72 xx 10^(-20)` ` :. r =20.7 xx 10^(-8) cm` Thus average distnace in between two `N_(2)` molecules ` = 2xx r = 41.4 xx 10^(-8) cm` Given radius of `N_(2) = 2 xx 10^(-8)cm` obtained radius is between two molecules is gaseous state is alomost 10 times of the diameter of each molecule This confirms the empty space in gaseous state and also a reasonable justification for their compression . |
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| 114. |
`20%N_(2)O_(4)` molecules are dissociated in a sample of gas at `27^(@)C` and 760 torr. Calculate the density of the equilibrium mixture. |
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Answer» `{:(,N_(2)O_(4)hArr,2NO_(2),,),("Mole presentinitially",1,0,,),("Mole at Eq",(1-0.2),0.4,,):}` `:.` M. wt. of mixture `= (0.8 xx 92 +0.4 xx 46)/(1.2) =76.66` Now `PV =(w)/(m)RT` or `(w)/(V) = (Pm)/(RT) = (76.66 xx 1)/(0.082 xx 300) = 3.116 g litre^(-1)` . |
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| 115. |
An iron cylider contains helium at a pressure of `250 kPa` at `300 K` The cylinder can withstand a pressure of `1 xx 10^(6)` Pa The room in which cylinder is placed catches fire Predict whether the cylinder will blow up before it melts or not melting point of cylinder `=1800 K` . |
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Answer» Given `P_(1) = 250 kPa, P_(2) = 1 xx 10^(6) Pa, T_(1) = 300K,` `T_(2) =?` Since volume of cylinder remains constant therefore `(P_(1))/(T_(1)) = (P_(2))/(T_(2))` `(250 xx 10^(3))/(300) = (1 xx 10^(6))/(T_(2))` `T_(2) = 1200K` The cylinder will blow upto `1200K` before its melting point `(1800K)` . |
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| 116. |
A x:1 molar mixture of He and `CH_(4)` is contained in a vessel as 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out.if the composition of the mixture effusing out initially is 8:1, then calculate the value of of x?A. 1B. 4C. 8D. 9 |
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Answer» Correct Answer - b |
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| 117. |
Critical temperature of `H_(2)O, NH_(3), CO_(2)` and `O_(2)` are 647 K, 405.6 K, 304.10 K and 1542 K respectively. If the cooling starts from 500 K to their critical temperature, the gas that lilquiefies first isA. `H_(2)O`B. `NH_(3)`C. `CO_(2)`D. `O_(2)` |
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Answer» Correct Answer - B Since the cooling starts from 500K, so water is not under consideration . As it is much easier to achieve 500K from 405.6 K. So, `NH_(3)` will get liquefied first. |
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| 118. |
If pressure of a gas contained in a closed vessel is increased by `0.4%` when heated by `1^(@)C`, the initial temperature must beA. 250KB. `250^(@)C`C. `25^(@)C`D. 25K |
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Answer» Correct Answer - a |
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| 119. |
A closed vessel at temperature T contains a mixture of two diatomic gases A and B. Atomic mass of A is 16 times that of B and mass of gas A contained in the vessel is 2 times that of B. which of the following statements are correct?A. Average kinetic energy per molecule of A is equal to that of B.B. Root mean square velocity of B four times that of A.C. Pressure exerted by B is eight times of that of A.D. Number of molecules of B, in cylinder, is eight times that A,. |
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Answer» Correct Answer - abcd |
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| 120. |
Two flasks of equal volume connected by a narrow tube (of negligible volume) are at `27^(@)C` and contain 0.70 mole of `H_(2)` to 0.5atm One of the flask is then immersed into a bath kept at `127^(@)C` while the other remains at `27^(@)C` Calculate the final pressure and the number of mole of `H_(2)` in each flask .A. Moles in flask 1-=0.4, Moles in flask 2=0.3B. Moles in flask 1=0.2,Moles in flask 2=0.5C. Moles in flask 1=0.3, Moles in flask 2=0.4D. Moles in flask 1=0.4, Moles in flask 2=0.2 |
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Answer» Correct Answer - a |
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| 121. |
Two flasks of equal volume connected by a narrow tube (of negligible volume) are at `27^(@)C` and contain 0.70 mole of `H_(2)` to 0.5atm One of the flask is then immersed into a bath kept at `127^(@)C` while the other remains at `27^(@)C` Calculate the final pressure and the number of mole of `H_(2)` in each flask .A. 0.5714 atmB. 1.5714 atmC. 0.5824atmD. none of the above |
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Answer» Correct Answer - a |
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| 122. |
Two flasks of equal volume is connected by a narrow tube (of negligible volume) contains a certain amount of `N_(2)` gas at 2 atm and `27^(@)C`. The 1st flasks is then immersed into a bath kept at `47^(@)C` while the 2nd flask is immeresed into a bath kept at `127^(@)C`. the ratio of the number of moles of `N_(2)` in 1st flask to the 2nd falsk after sometime will be?A. `5:4`B. `2:3`C. `3:2`D. `4:5` |
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Answer» Correct Answer - a |
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| 123. |
A closed container of volume `0.02m^(3)` contains a mixture of neon and argon gases at a temperature `27^(@)C` and pressure `1 xx 10^(5) Nm ^(-1)` The total mass is 28 and the molar mass of and argon are `20` and `40` respectively find the masses of individual gases in the container assuming then to be ideal . |
| Answer» Correct Answer - Neon =4g, Argon =24 g | |
| 124. |
Three hollow metallic boxes `A,B` and `C` are connected to one another through narrow tube of negligible volume and are filled with argon gas if the internal volume of these boxes are in the ratio 1:2:4 find out The molar ratio of orgon in these boxes (b) The mole ratio of argon if boxes `A,B` and `C` are immersed in differnet temperatures bath having `27^(@)C,127^(@)C` and `327^(@)C` respectively . |
| Answer» Correct Answer - 1:2:4, (b) 2:3:4 | |
| 125. |
An iron meteorite was analysed for its argon content The amount of Ar was `0.20mm^(3)` `(STP)` per kg of meteorite. If each Ar atom had been formed by a single cosmic event how many such events must there have been per kg of meteorite ? . |
| Answer» Correct Answer - `5.4 xx10^(150 atoms` | |
| 126. |
The figure shows initial conditions of a uniform cylinder with frictionless pistons A and B held in shown position by mechanical stoppers. If the mechanical stoppers holding piston A and B as shown in figure are removed and in the mean time `N_(2)O_(4) and O_(3)` gases separately undergo following reaction completely [Assume that temperature remains constant] `N_(2)O_(4) rarr 2NO_(2)` `2O_(3) rarr 3O_(2)` Which statement is correct after all the pistons have attained their final positions (assume `sigma` to be same for all gases)?A. `Z_(11)` will be highest for molecules of gas in compartment 1B. Mean free path will be longest for molecules of gas in compartments2C. `Z_(11)` will be same for all gasesD. Mean free path will be same for molecules of gas in all compartments |
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Answer» Correct Answer - d |
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| 127. |
Addition of surface to water leads .A. An increase in angle of contactB. An increase in surface tensionC. A decrease in angle of contact and increase in surface tension .D. A decrease in angle of contact and increase in surface tension . |
| Answer» Addition of soap redues surface tension of system. | |
| 128. |
`SO_(2)` at STP contained in a flask was replaced by `O_(2)` under identical conditions of pressure, temperature and volume. Then the weight of `O_(2)` will be.... Of `SO_(2)`A. halfB. one fourthC. twiceD. four times |
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Answer» Correct Answer - a |
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| 129. |
Which noble gas effuses approximately twice are fast as Kr? A. NeB. ArC. XeD. Rn |
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Answer» Correct Answer - a |
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| 130. |
An unknown gas effuses through a small hole one half as fast as methane, `CH_(4)`, under the same conditions. What is the molar mass of the unknown gas?A. `4g "mol"^(-1)`B. `8g "mol"^(-1)`C. `32g "mol"^(-1)`D. `64g. "mol"^(-1)` |
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Answer» Correct Answer - d |
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| 131. |
Gas molecules each of mass `10^(-26)` kg are taken in a container of volume `1 dm^(3)` . The root mean square speed of gas molecules is 1 km `sec^(-1)` .What is the temperature fo gas molecules. (Given : `N_(A) =6xx10^(23),R=8J//mol.K`)A. 298 KB. 25 KC. 250 KD. 2500 K |
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Answer» Correct Answer - c (c) `1000=sqrt((3RT)/(M)),10^(3)=[(3xx8.314xxT)/(10^(-26)xx6xx10^(23))]^(1//2)` ` T=(10^(6)xx10^(-26)xx6xx10^(23))/(3xx8)=250K` |
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| 132. |
Two glass bulbs A (of 100 mL capacity), and B (of 150mL capacity) containing same gas are connected by a small tube of negligible volume. At particular temperature the pressure to A was found to be 20 times more than that in bulb B. the stopcock is opened without changing the temperature. the pressure in A will:A. drop by 75%B. drop by 57%C. drop by 25%D. will remain same |
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Answer» Correct Answer - b |
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| 133. |
Rate of diffusion of 1 mole `CO` and 2 mole `N_(2)` in a container are same Rate of diffusion `prop sqrt(1//M)` at constant `P,T` . |
| Answer» `r prop (P)/(sqrt(M))` at constant `T` . | |
| 134. |
A mixture contains `N_(2)O_(4)` and `NO_(2)` in the ratio 2:1 by volume. The vapour density of the mixture is:A. 45.4B. 49.8C. 32.6D. 38.3 |
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Answer» Correct Answer - d |
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| 135. |
The constant motion and high velocities of gas particles lead to some important practical consequences One such consequence is that is mixing rapidly when they come in contact. The mixing of different gases by random molecular motion and with frequent collisions is called diffusion.A similar process in which gas molecules escape through a tiny hole into vaccum is called effusion. Helium gas at 1 atm and `SO_(2)` at 2 atm pressure temperature being the same are released seperately at the same moment into 1 m long evacuated tubes of equal diameters If helium reaches the other end of the tube in t sec what distance `SO_(2)` would traverse in the same time interval in the other tube?A. 25 cmB. 50 cmC. 60 cmD. 75 cm |
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Answer» Correct Answer - b |
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| 136. |
An ideal gas can never be liquefied because .A. Its critical temperature is always above `0^(@)C` .B. Its molecules are relatively small in sizeC. In solidifies before becoming liquidD. Forces of attractions among ideal gas molecules are negligible . |
| Answer» To get a liquid appeciable value for forces of attractions among molecules is must . | |
| 137. |
A liquid is in equilibrium with its vapour at its boiling point. On average, the molecules in the two phases have equalA. Intermolecular forcesB. Kinetic energyC. Total energyD. Potential energy |
| Answer» `K.E` of gas and liquid molecules at may temperature is given by `(3)/(R )` At equilibrium both have same `T` and thus same `K.E` . | |
| 138. |
Select correct statements:A. A real gas can be liquified at critical temperatureB. Critical pressure is the maximum pressure at which a substance is present in its liquid state at `T_(C)`C. Ideal gas can be liquified below `T_(C)`D. Critical volume is the molar volume of substance in gaseous state at `T_(C)` and `P_(C)` |
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Answer» Correct Answer - a,d (a,d) Critical pressure is the minimum pressure at which substance convert into liquid state at critical point. |
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| 139. |
Which is/are correct for real gases?A. `underset(Prarr0)(Lim)(PV_(m))=` constant at constant high temperatureB. `underset(V_(m)rarr0)(Lim)(PV_(m))=` constant at constant low temperatureC. `underset(Prarr0)(Lim)((PV_(m))/(RT))=1` at high temperatureD. `underset(Vrarr0)(Lim)((PV_(m))/(RT))=R` |
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Answer» Correct Answer - a,c (a,c) Low P and high temperature gas henaves as an ideal gas. `:. PV=`constant and `(PV_(m))/(RT)=1.` |
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| 140. |
If a gas expands at constnat temperature .A. The kintetic energy of molecules remains sameB. The kinetic energy of molecules increasesC. The number of molecules of gas increasesD. The pressure increases |
| Answer» Correct Answer - `K.E prop T` . | |
| 141. |
When hydrogen gas at high pressure and room temperature expands adiabatically to a region of low pressure there is a decreases in temperature Hydrogen gas at room temperature is above its inversion temperature . |
| Answer» Hydrogen gas at high pressure and room temperature expands adiabatically into a region of low pressure there is a increase in temperature . | |
| 142. |
Comment about the fraction of molecules moving between 400 to 500m/sec for a gas (molecular mass =20 g/mol) if its temperature increases from 300K to 400K [25/3 J/mol/K].A. Fraction of molecules increasesB. Fraction of molecules decreasesC. Fraction of molecules remains constantD. Fraction of molecuels remains constant |
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Answer» Correct Answer - b |
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| 143. |
The average speed at temperature `T^(@)C` of `CH_(4)(g)` is `sqrt((28)/(88))xx10^(3)ms^(-1)`. What is the value of T ?A. `240.55^(@)C`B. `-32.45^(@)C`C. `3000^(@)C`D. `-24.055^(@)C` |
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Answer» Correct Answer - b (b) `sqrt((8RT)/(piM))=sqrt((28)/(88))=sqrt((7)/(22))` `(8xx8.314xxT)/(pixx16xx10^(-3))=(7)/(22)xx10^(6)` `T=(1000xx2)/(8.314)=240.55K` `T^(@)C=240.55-273=-32.45^(@)C` |
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| 144. |
The graph of compressibility factor (Z) vs. P for one mole of a real gas is shown in following diagram. The graph is plotted at constant temperature 273K. If the slope of graph at very high pressure `((dZ)/(dP))` is `((1)/(2.8))atm^(-1)` , then calculate volume of one mole of real gas molecules (in L/mol) Given : `N_(A)=6xx10^(23)` and `R=(22.4)/(273)L atmK^(-1)mol^(-1)` |
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Answer» Correct Answer - 2 `Z=1+(Pb)/(RT)" high pressure"` `(dZ)/(dP)=(b)/(RT)=(1)/(2.8)` `b=(RT)/(2.8)=(22.4)/(2.8)=4xx(N_(A)xx(4)/(3)piR^(3))` `(N_(A)xx(4)/(3)piR^(3))="Volume of 1 mole gas"` `=(5.6)/(2.8)=2` |
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| 145. |
A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300K. The ratio of the average rorational kinetic energy per `O_2` molecules to that per `N_2` molecules isA. `1:1`B. `1:2`C. `2:1`D. Depends upon momnets of inertia of two molecules . |
| Answer» Both are diatomic gases and thus possess two degree of freedom associated with rotational kinetic energy. At same `T` they have same kinetic energy . | |
| 146. |
The kinetic energy of two moles of `CO_(3)` at a certain temperature is 1800cal. The temperature of the gas is :A. 300KB. 150KC. 200KD. 400K |
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Answer» Correct Answer - a |
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| 147. |
A 10 g sample of oxygen gas is taken in a container of volume 1 litre and is found to exert a pressure of 3 bar. Which of the following options is correct regarding speed of the molecuels?A. All the molecules are moving at a same speed which is equal to 310 m/sec.B. `U_("avg")=300m//"sec"`C. `U_("mps")=300xxsqrt((2)/(5)` m/sec.D. `U_("mps")=310` m/sec. |
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Answer» Correct Answer - c |
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| 148. |
Oxygen gas generated by the decomposition of potassium chlorate is collected. The volume of oxygen collected at `24^(@)C` and atmospheric pressure of `760 m m Hg` is `128 mL`. Calculate the mass `(` in grams `)` of oxygen gas obtained. The pressure of water vapour at `24^(@)C` is `22.4 m m Hg`.A. 1.36 gB. 1.52 gC. 0.163 gD. 1.63 g |
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Answer» Correct Answer - c (c) Form the total pressure and the vapour pressure of water we can calculate the partial pressure of `O_(2)`. `P_(O_(2))=P_(T)-P_(H_(2)O)` `=760-22.4` `=737.6 mm " "Hg` From the ideal gas equation we write `m=(PVM)/(RT)` `=((0.974atm)(0.128L)(32.0g//mol))/((0.0821Latm//K mol)(273+24)K)` `=0.163g` |
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| 149. |
The mass of molecule `A` is twice the mass of molecule B. The rms speed of A is twice the rms speed of B. If two samples of A and B contain same number of molecules. If the pressure of gas B is 2atm then what will be the pressure of gas A ( atm) . If two samplest are taken in separate containers of equal volume ?A. 16B. 32C. 48D. 64 |
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Answer» Correct Answer - 1 Given `m_(A)=2m` `:." "`mol. `wt. ` of `A=2xxmol." " wt." "` of `B" "….(i)` Given`" "U_(rms)` of `A=2xxu_(rms)` of `B " " ….(ii)` Also number of molecules of `A =` number of molecules of `B" "…..(iii)` For gas `A" "P_(A)V_(A)=(1)/(3)M_(A)u_(rmsA)^(2)` For gas `B" "P_(B)V_(B)=(1)/(3)M_(B)u_(rmsB)^(2)` `:. " "(P_(A)V_(A))/(P_(B)V_(B))=(M_(A))/(M_(B))xx(u_(A)^(2))/(u_(B)^(2))" "....(iv)` Given `" "V_(A)=V_(B)" "....(v)` `:." "` By equation `(i),(ii),(iv)` and `(v)` `(P_(A))/(P_(B))=2xx(2)^(2)=8` `:." "P_(A)=8P_(B)` Ans. 16 |
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| 150. |
Which of the following samples of ideal gas will have maximum translation kinetic energy??A. `(1)/(2)` mole of `CO(g)` at 400KB. 16g of oxygen gas at 200KC. 28g of nitrogen gas at 300KD. 1g of ozone at 600K |
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Answer» Correct Answer - c |
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