A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at…

A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure P0, volume V0 and temperature T0. If the gas mixture is adiabatically compressed to a volume V04, then the correct statement(s) is/are,
(Given 21.2= 2.3; 23.2=9.2; R is gas constant)
  1. The final pressure of the gas mixture after compression is in between 9P0 and 10P0.
  2. The average kinetic energy of the gas mixture after compression is in between 18RT0 and 19RT0.
  3. The work W done during the process is 13RT0.
  4. Adiabatic constant of the gas mixture is 1.6.

Solution

Here, n1=5,CP1=52R,CV1=32Rn2=1, CP2=72R,CV2=52Rγmix=n1CP1+n2CP2n1CV1+n2CV2=85=1.6
Work done, W=P1V1-P2V2γ-1, In adiabatic process
Now, for adiabatic compression
P1V1r=P2V2r
⇒P0V08/5=P2V048/5
⇒P2=9.2P0⇒A is correct.
Now, W=P0V0-9.2P0V043/5=-13RT0⇒C is correct
∴W=13 RT0
Again T1V1γ-1=T2V2γ-1
T2=T126/5=2.3T0
Average kinetic energy of gas mixture =nCVmixT2
=23RT0 ⇒B is incorrect

Asked in: JEE Advanced 2019 (Paper 2)

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