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Kinetic Theory · Question 21

AIPMT 2015MCQMedium📈 Topic in 12% papers+4−1
1

4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is 5.0 J K-1 mol-1. If the speed of sound in this gas at NTP is 952 m s-1, then the heat capacity at constant pressure is (Take gas constant R = 8.3 J K-1 mol-1)

  1. 7.0 J K-1 mol-1
  2. 8.5 J K-1 mol-1
  3. 8.0 J K-1 mol-1
  4. 7.5 J K-1 mol-1

Answer

(3) 8.0 J K-1 mol-1

Solution

22.4 L at NTP is 1 mol M = 4.0 g mol-1 = 4 10-3 kg mol-1

v = RTM = v2MRT

= (952)2 4 10-38.3 273

= 3625.22265.9 1.6

Cp = Cv = 1.6 5.0 = 8.0 J K-1 mol-1

Answer: (3)

AIPMT 2015

Keep practising — Kinetic Theory

2

The ratio of the specific heats CpCv = in terms of degrees of freedom (n) is given by

  1. (1 + 2n )
  2. (1 + n2 )
  3. (1 + 1n )
  4. (1 + n3 )
AIPMT 2015 (Cancelled)
3

The mean free path of molecules of a gas, (radius r) is inversely proportional to

  1. r3
  2. r2
  3. r
  4. r
AIPMT 2014
4

In the given (V - T) diagram, what is the relation between pressure P1 and P2? [figure]

  1. P2 < P1
  2. Cannot be predicted
  3. P2 = P1
  4. P2 > P1
NEET 2013
5

The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from T1 K to T2 K is

  1. 34Na kB (T2 - T1)
  2. 34Na kB ( T2T1 )
  3. 38Na kB (T2 - T1)
  4. 32Na kB (T2 - T1)
NEET 2013
6

In a vessel, the gas is at pressure P. If the mass of all the molecules is halved and their speed is doubled, then the resultant pressure will be

  1. 2P
  2. P
  3. P/2
  4. 4P
NEET 2013 (Karnataka)
7

At 10 the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110 this ratio is

  1. 10110x
  2. 283383x
  3. x
  4. 383283x
AIPMT 2008
8

The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be (where R is the gas constant)

  1. PV = (5/32)RT
  2. PV = 5RT
  3. PV = (5/2)RT
  4. PV = (5/16)RT
AIPMT 2004
9

To find out degrees of freedom, the expression is

  1. f = 2 - 1
  2. f = + 12
  3. f = 2 + 1
  4. f = 1 + 1
AIPMT 2000

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