JEE Main 1-Sep-2021 Shift 2 Solved Paper

Given, the number of diatomic moles, n = 3 mol
The increase in temperature of the diatomic mole,
ΔT = 40ºC
Now, the degree of freedom,
f = linear + rotational + no oscillation
f = 3 + 2 + 0
f = 5
Change in internal energy,
ΔU=nCVΔT
where,
CV=

f

2

R=

5

2

R
Substituting the value in Eq. (i), we get
ΔU=

5R

2

nΔT
Now, work done by the gas for isobaric process,
W=pΔV=nRΔT
The ratio of the change in internal energy to the work done by thegas,

ΔU

W

=

5 2 nRΔT

nRΔT

ΔU

W

=

5

2

Multiply and divide the above equation with 5, we get

ΔU

W

=

5×5

2×5

=

25

10

Comparing with given equation,

ΔU

W

=

x

10

The value of the x = 25.