NEET
Practice Test 12
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Two flasks A and B of equal volumes maintained at temperatures 300K and 600K contain equal mass of H2 and CH4 respectively. The ratio of total translational kinetic energy of gas in flask A to that in flask B is

  • Solution

A mixture of two gases A and B in the mole ratio 2 : 3 is kept in a 2 litre vessel. A second 3L vessel has the same two gases in the mole ratio 3 : 5. Both gas mixtures have the same temperature and same total pressure. They are allowed to inter-mix and the final temperature and the total pressure are the same as the initial values, the final volume being 5 litres. Given that the molar masses are MA and MB. What is the mean molar mass of the final mixture?

  • Solution

The pressure of 11gm of a gas which is placed in a 4 litres container at 127°C is 2 atm, then the gas would be–(Take : R = 0.08 litre atm K-1mol-1)

  • Solution

    Calculate molar mass M,

    \(M = \frac{mRT}{PV} = \frac{11 \times 0.08 \times 400}{2 \times 4} = 44\)

The van der Waals’s constants for gases A, B and C are as follows

GAS a(L2atm mol-2) b(L mol-1)
A 0.024 0.027
B 4.17 0.037
C 3.59 0.043

Based upon the above data, which of the following statements is correct?

(i) The gas B has the highest critical temperature
(ii) The gas A has minimum departure from the ideal behavior
(iii) The gas C has largest molecular volume

  • Solution

    (i) Greater is the van der Waals’s constant a, higher would be Tc(easier liquefaction of the gas).

    (ii) Smaller the constants a and b,lesser departure from ideal behaviour.

    (iii) Greater the constant b, larger is the molecular volume.

The compressibility factor Z = PV⁄RT for 1 mol of a real gas is greater than unity at a pressure of 1 atm and 273 .15 K. The molar volume of the gas at STP will be

  • Solution

    The gas is less compressible than ideal gas. Hence , Vm > 22.4L

There are three closed containers in which equal amount of the gas are filled.

If all the containers are placed at the same temperatures, then find the incorrect options –

  • Solution

    n, T same hence P ∝ 1⁄V,

    V1 = 1000 cm3

    V2 = π(10)2 × 10 = 1000πcm3

    V3 = 4⁄3π(10)3 = 4⁄3 π 1000 cm3

    ∴ Pressure of the gas is minimum in (III) container, pres-sure of the gas is maximum in (I),

    The ratio of pressure in II and III container is 4 : 3

At what temperature the average speed of helium molecule will be the same as that of oxygen molecule at 527°C

  • Solution

    For equal average speeds of two of gases,

    \(\sqrt{\frac{T_{1}}{M_{1}}} = \sqrt{\frac{T_{2}}{M_{2}}}\)

    T1 = ?; M1 = 4; T2 = 273 + 527 = 800K; M2 = 32

Positive deviation from ideal behaviour takes place because of

  • Solution

    For positive deviation: PV = nRT + nPb

    ⇒ \(\frac{PV}{nRT} = 1 + \frac{Pb}{RT}\)

    Thus, the factor nPb is responsible for increasing the PV value, above ideal value. b is actually the effective volume of molecules. So, it is the finite size of molecules that leads to the origin of band hence positive deviation at high pressure.

If for two gases of molecular masses MA and MB at temperatures TA and TB, TA MB = TB MA, then which of the following properties has the same magnitude for both the gases

Which of the following volume (V) – temperature (T) plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure ?

  • Solution

    Find the volume by either V = RT/P (PV = RT) or P1V1 = P2V2 and match it with the values given in graph to find correct answer.

    Volume of 1 mole of an ideal gas at 273 K and 1 atm pressure is 22.4 L and that at 373 K and 1 atm pressure is calculated as ;

    \(V = \frac{RT}{P} = \frac{0.082 \times 373}{1}\) = 30.58 L = 30.6 L

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