InterviewSolution
Saved Bookmarks
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.
| 501. |
At low pressures, the van der Waals equation is written as `[P+(a)/(V^(2))]V=RT` The compressibility factor is then equal toA. `(1-(a)/(RTV))`B. `(1-(RTV)/(a))`C. `(1+(a)/(RTV))`D. `(1+(RTV)/(a))` |
|
Answer» Correct Answer - a (a) `(PV)/(RT)=Z " ":. Z=1-(a)/(VRT)` |
|
| 502. |
Assertion`:` Compressiblity factor for hydrogen varies with pressure with positive slope at all pressures. Reason `:` Even at low pressure, repulsive forces dominate hydrogen gas.A. If both assertion and reason are true and the reason is correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false |
|
Answer» Correct Answer - A In case of `H_(2)`, compressibility factor increase with the pressure . At 273 K, `Z gt 1` which shows that it is difficult to compress the gas as compared to ideal gas. In this case repulsive forces dominate. |
|
| 503. |
The volume temperature graph of a given mass of an ideal gas at constant pressure are shown below What is the correct order of pressuresA. `P_(1) gt P_(3) gt P_(2)`B. `P_(1) gt P_(2) gt P_(3)`C. `P_(2) gt P_(3) gt P_(1)`D. `P_(2) gt P_(1) gt P_(3)` |
|
Answer» Correct Answer - A The correct order of pressure is `P_(1) gt P_(3) gt P_(2)` |
|
| 504. |
Equal volume of two gases which do not react together are enclosed in separate vessels. Their pressures are 10mm and 400mm respectively. If the two vessels are joined together, then what will be the pressure of the resulting mixture (temperature remaining constant)?A. 120mmB. 500mmC. 1000mmD. 205mm |
|
Answer» Correct Answer - d |
|
| 505. |
A bulb of constant volume is attached to a manometer tube open at other end as shown in figure. The manometer is filled with a liquid of density `(1//3)^("rd")` that of mercury. Initially h was 228 cm. Through a small hole in the bulb gas leaked assuming pressure decreases as `(dP)/(dt)=-kp.`If value of h is 114 cm after 14 minutes, what is the value of k (in `"hour"^(-1)`)? [ Use : In (4/3) =0.28 and density of Hg =13.6 g/mL.].A. 0.6B. 1.2C. 2.4D. none of these |
|
Answer» Correct Answer - b (b) `P=P_(0^(e^(-kt))),P_(0)=(228)/(3)+76=152cm Hg` `"At "t=14 min,P=(114)/(3)+76=114cm Hg` `implies" "152e^(-kt)=114` `impliesk=(1)/(t)"In"(152)/(114)=(1)/(14)xx0.28` `=0.02xx60hr^(-1)=1.2hr^(-1)` |
|
| 506. |
When iodine is heated under atmospheric pressure conditions it transforms to vapours without passing through the liquid state . If he triple point pressure of a system is high and unless external pressure is applied to exceed the triple-point pressure sublimation will take place . |
| Answer» Explantion is correct reason for statement | |
| 507. |
If density of vapours of a substance of molar mass `18 gm at 1 atm` pressure and`500K` is `0.36 kg m^(-3)`, then calculate the value of `Z` for the vapours. (Take `R = 0.082 L atm mole^(-1)K^(-1))` |
| Answer» Correct Answer - `(50)/(41)` | |
| 508. |
Let `P` and `P_(s)` be the partial pressure of water and saturated partial pressure of water vapours, then relative humidity is given by .A. `(P_(s)+P)/(P_(s)) xx100`B. `(P)/(P_(s)) xx100`C. `P_(s)/(P) xx 100`D. `(P+P_(s)) xx 100` |
|
Answer» Relative humidity `("Vapour pressure of water "xx100)/("Saturated vapour pressure")` . |
|
| 509. |
A gas `A_(x)` having mass of 100g is confined in a container of volume 16L maintained at 300K and a manometer is attached as shown, value of x is: (R=0.08 atm-L/mole-K) (Atomic mass of A=24) A. 2B. 8C. 4D. 5 |
|
Answer» Correct Answer - d |
|
| 510. |
Statement-1: The total kinetic energy of vapours formed over liquid `H_(2)O(l)` in two closed container A and B having free space 1 litre and 2 litre respectively over `H_(2)O(l)` at the same temperature is in the ratio 1:2. (Assuming ideal gas behaviour of vapours). Statement-2: Vapour pressure of a substance depends only on temperature.A. Statement-I is True, Statement-II is True : Statement-II is a correct explanation for Statement-IB. Statement-I is True, Statement-II is True : Statement-II is `NOT` a correct explanation for Statement-IC. Statement-I is True, Statement-II is False.D. Statement-I is False, Statement-II is True. |
|
Answer» Correct Answer - a |
|
| 511. |
Select the correct observartion for a 8.21 litre container, fillled with 2 moles of He at 300K/A. It has pressure 6 atm.B. If it is an open rigid container, its pressure increases to 8 atm on heating to 400KC. IF it is closed non-rigid (like thin skin balloon), its volume increases to 16.42 lit on heating to 600K.D. When connected with another similar empty container maintained at 150K while maintaining original container at 300 K, pressure reduced to `(2)/(3)` atm. |
|
Answer» Correct Answer - ac |
|
| 512. |
If 10 moles of a real gas is present in 2 litre container having a free volume of 1600 mL at pressure P and temperature T, then at what pressure in atmosphere and at 273 K its compressibility factor is 1.5 ? (neglect `alpha`) |
|
Answer» Correct Answer - 280 |
|
| 513. |
From the following set-up, calculate moles of the gas present in the container of volume 24.63 litre at 600K if the level of mercury in the open tube of the manometer is 36cm higher. [Given: Atmospheric pressure=78cm of mercury] A. 1.5 molesB. 0.73 molesC. 3 molesD. 1 mole |
|
Answer» Correct Answer - b |
|
| 514. |
One mole of `N_(2)O(g)` at `300K` is kept in a closed container under one atmosphere. It is heated to `600K` when `20%` by mass of `N_(2)O_(4)(g)` decomposes of `NO_(2)(g)`. The resultant pressureA. `1.2 atm`B. 2.4 atmC. `2.0` atmD. `1.0` atm |
|
Answer» Correct Answer - B `N_(2)O_(4)(g) hArr 2NO_(2)(g)` At start `=100 // 92 mol = 1.08 mol ` 0 At equilibrium `80 // 92 mol = 0.86` `20 // 46 mol = 0.43 mol` According to ideal gas equation, at two condition At 300 K `P_(0)V=n_(0)RT_(0)` `1 xx V = 1.08 xx R xx300` ....(i ) At 600K `P_(1)V = n_(2)RT_(1)` `P_(1)xxV = (0.86+0.43) xx R xx 600` ....(ii) Divide (ii) by (i) `(P_(1))/(1) = (1.29xx600)/(1.08xx300), P_(1)= (1.29xx2)/(1.08) = 2.38 atm`. |
|
| 515. |
The root mean square velocity of an ideal gas at constant pressure varies with density (d) as:A. `d^(2)`B. `d`C. `sqrt(d)`D. `(1)/sqrt(d)` |
|
Answer» Correct Answer - d |
|
| 516. |
Which of the following has maximum root mean square velocity at the same temperature ?A. `SO_(2)`B. `CO_(2)`C. `O_(2)`D. `H_(2)` |
|
Answer» Correct Answer - D `u_(rms)=sqrt((3RT)/(m)):. u_(rms) prop (1)/(sqrt(M))` at same T because `H_(2)` has least molecular weight so its r.m.s. velocity should be maximum. |
|
| 517. |
In the temperature changes from `27^(@)C` to `127^(@)C`, the relative percentage change in RMS velocity isA. 1.56B. 2.56C. 15.6D. 82.4 |
|
Answer» Correct Answer - C At `27^(@)C, C_(rms) ` will be `sqrt(300) x - 17.3 x` `(x= sqrt((3R)/(M)))` At `127^(@)C `, it is `sqrt(400)x = 20 x` Percentage increase ` = ((20-17.3))/(17.3x)xx 100 = 15.6` |
|
| 518. |
The ratio of most probable velocity to that of average velocity isA. `pi//2`B. `2//pi`C. `sqrt(pi)//2`D. `2 // sqrt(pi)` |
|
Answer» Correct Answer - C `(alpha_(mp))/(v_(av))=(sqrt((2RT)/(M)))/(sqrt((8RT)/(piM)))=(sqrt(pi))/(2)` |
|
| 519. |
The ratio of most probable velocity to that of average velocity isA. `1.128`B. `1.224`C. `1.0`D. `1.112` |
|
Answer» Correct Answer - A Average speed `:` most probable speed `sqrt((8RT)/(piM)): sqrt((2RT)/(M))implies sqrt((8)/(pi)):sqrt(2)implies 1.128 :1`. |
|
| 520. |
At what temperature, the r.m.s. velocity of a gas measured at `50^(@)C` will become double ?A. 626KB. 1019KC. `200^(@)C`D. `1019^(@)C` |
|
Answer» Correct Answer - D `C _(rms)=sqrt((3RT)/(M))` Where T is in kelvin scale. For rms velocity to be doubled, T in kelvin scale must be raised 4 times. So it should be `4 xx (50 +273) =1292K = 12920273=1019^(@)C`. |
|
| 521. |
A vessel of 10 L capacity contains 4 g He gas. The vessel is heated such that its absolute temperature becomes double. In order to make the pressure of gas half of its initial pressure, the mass of gas (in g) which should be removed from the vessel is : (Assume no change in the capacity of vessel) : |
|
Answer» Correct Answer - `0003` |
|
| 522. |
Value of absolute zero of temperature in degree Celsius `(.^(@)C)` can be determined by given data. The density of an ideal gas at `25^(@)C` and `100^(@)C` are 1.5 and 1.2g/L, respectively, both at the same pressure. The value of absolute zero of temperature in degree Celsius `(.^(@)C)`:A. `-273^(@)C`B. `-275^(@)C`C. `-200^(@)C`D. `-0^(@)C` |
|
Answer» Correct Answer - b |
|
| 523. |
If the absolute temperature of an ideal gas become double and pressure become half , the volume of gas would beA. Remain unchangeB. Will be doubleC. Will be four timesD. Will be half |
|
Answer» Correct Answer - C `P(1)=P,V_(1)=V,T_(1)=T` `P_(2)=(P)/(2),V_(2)=? , T_(2)=2T` According to gas equation `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`or `(PV)/(T)=(PV_(2))/(4T)implies V_(2)=4V` |
|
| 524. |
Atmolysis is a process ofA. Atomising gas moleculesB. The breaking of atoms to sub-atomic particlesC. Separation of gases from their gaseous mixtureD. Changing of liquids to their vapour state |
|
Answer» Correct Answer - C Gases may be separated by this process because of different rates of diffusion due to difference in their densities. |
|
| 525. |
A bottle of dry `NH_(3)` and another bottle of dry `HCl` connected through a long tube are opened simultaneously at both ends of the tube. The white ring `(NH_(4)Cl)` first formed will beA. AB. BC. CD. A,B and C simultaneously |
|
Answer» Correct Answer - c |
|
| 526. |
A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will beA. At the centre of the tubeB. Near the hydrogen chloride bottleC. Near the ammonia bottleD. Throughout the length of the tube |
|
Answer» Correct Answer - B Rate of diffusion `prop sqrt(1// "Mol. Mass")` . Mol. Mass of `HCl gt` Mol. Mass of `NH_(3) NH_(4)Cl` ring will first formed near the HCl bottle because rate of diffusion of `NH_(3)` is more than that of HCl ( because `M_(NH_(3)) : M_(HCl)=17:36.5)`. SO `NH_(3)` will reach first to the HCl bottle & will react there with HCl to form `NH_(4) Cl` ring |
|
| 527. |
What conclusion would you draw from the following graphs for an ideal gas? A. As the temperature is reduced, the volume as well as the pressure increaseB. As the temperature is reduced, the volumebecomes zero and the pressure reaches infinityC. As the temperature is reduced, the pressure decreaseD. A point is reached where, theoretically, the volume become zero |
|
Answer» Correct Answer - c,d (c,d) `{: ("at constant volume","at constant P"),(PpropT,VpropT),(P=KT,V=KT),(P=K(t^(@)C+273.15),V=K(t^(@)C+273.15)):}` |
|
| 528. |
A certen amount of gas at `2.5^(@)C` and at a pressure of 0.80 atm is kept in a glass vessel. Suppose that the vessel can withstand a pressure of 2.0 atm. How high can you raise the temperature of the gas without bursting the vessel?A. `745^(@)C`B. `472^(@)C`C. `500^(@)C`D. None of these |
|
Answer» Correct Answer - b (b) `(P_(1))/(T_(1))=(P_(2))/(T_(2))` |
|
| 529. |
A teacher enters a classroom from front door while a student from back door. There are 13 equidistant rows of benches in the classroom. The teacher releases `N_(2)O`, the laughing gas, from the first bench while the student releases the weeping gas `(C_(6)H_(11)OBr)` from the last bench. At which row will the students starts laughing and weeping simultaneously?A. 7B. 10C. 9D. 8 |
|
Answer» Correct Answer - c |
|