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 V0/4, then the correct statement(s) is/are,
(Given 21.2=2.3;23.2=9.2;R is gas constant)
The work |W| done during the process is 13RT0
n1=5 moles CV1=3R2
n2=1 mole CV2=5R2
(CV)m=n1CV1+n2CV2n1+n2=5×3R2+1×5R26=5R3
γn=(Cp)m(Cv)m=85
So, option (3) is correct.
P0V8/50=P2(V04)8/5
P2=9.2P0
Option 'A' is also correct.
(2) T1Vγ−11=T2Vγ−12
T2=T1(2)6/5=2.3T0
Average kinetic energy of gas mixture = nCVmixT2 =23RT0
(4) W=P1V1−P2V2γ−1
W=P0V0−9.2P0V043/5=−13RT0
∴|W|=13RT0