Q.

A mixture of ideal gas containing 5 moles of monoatomic 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)           [2019]

1 The work |W| done during the process is 13RT0.  
2 The final pressure of the gas mixture after compression is in between 9P0 and 10P0.  
3 The average kinetic energy of the gas mixture after compression is in between 18RT0 and 19RT0.  
4 Adiabatic constant of the gas mixture is 1.6.  

Ans.

(1, 2, 4)

Adiabatic constant of the gas mixture,

γm=n1Cp1+n2Cp2n1Cv1+n2Cv2=5×5R2+1×7R25×3R2+1×5R2=1.6

For an adiabatic process, PVγ=Constant

  P=P0(V0V)1.6=P0(4)1.6

=P0(22)1.6=P023.2=9.2P0

Work done during the process,

W=P2V2-P1V11-γ=9.2P0×(V04)-P0V01-1.6=-13P0V06

But P0V0=6RT0    (as n=5+1=6)

 W=-13(6RT0)6=-13RT0

 |W|=13RT0

The average K.E. of the gas mixture,

K.E.¯=nCVmi×T2

From, T1V1γ-1=T2V2γ-1

or, T2=T1(2)6/5=23T0

 K.E.¯=nCVmi×T2=23RT0