JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, number of mole, n=1
Heat supplied =Q
Work done =Q∕5
By using first law of thermodynamics,
∆Q‌‌=∆U+∆W
Q‌‌=∆U+Q∕5
∆U‌‌=4Q∕5=nCv∆T. . . (i)
where, CV is heat coefficient at constant volume and
∆T is change in temperature.
Degree of freedom of diatomic gas, f=5
∴‌‌Q=‌

5

4

‌

f

2

R∆T [from Eq. (i)] ⇒‌‌‌

Q

∆T

=C=‌

5

4

×‌

5

2

R=‌

25

8

R
Hence, x=25