JEE Advanced 2020 Full Test 2 Paper 2

Let,
CV= molar heat capacity of the first gas,
CV′′= molar heat capacity of the second gas,
CV= molar heat capacity of the mixture,
and similar symbols for other quantities. Then
γ=‌

CP′

CV′

=1.67
and
C′P=CV′+R
This gives
CV′=3∕2R and CP′=5∕2R
Similarly, γ=1.4 gives CV′′=5∕2R and CP′′=7∕2R
Suppose the temperature of the mixture is increased by dT. The increase in the internal energy of the first gas =n1CV 'dT.
The increase in internal energy of the second gas =n2CV′′dT
Thus,
(n1+n2)CVdT=n1CV⋅dT+n2CV′′dT
CV‌‌=‌

n1CV′+n2CV′′

n1+n2

CP‌‌=CV+R=‌

n1CV′+n2CV′′

n1+n2

+R
‌‌=‌

n1(CV′+R)+n2(CV′′+R)

n1+n2

‌‌=‌

n1CP′+n2Cp′′

n1+n2

..........(2)
From (1) and (2)
γ‌‌=‌

CP

CV

=‌

n1CP′+n2CP′′

n1CV′+n2CV′′

=‌‌‌

4× 5 2 R+2× 7 2 R

4×‌ 3 2 R+2×‌ 5 2 R

=1.54