JEE Main 26 Feb 2021 Shift 2 Solved Paper

Given, ‌‌V=kT2∕3‌‌... (i)
where, T is temperature.
Change in temperature, ∆T=90K
Let p be the pressure, dV be the change in volume and work done be W.
As we know that,
W=∫pdV‌‌...‌ (ii) ‌
As, ‌‌pV=nRT
∴‌‌p=‌

nRT

V

Substituting this value in Eq. (ii), we get
‌W=∫nRT‌

dV

V

‌W=∫nRT‌

dV

kT2∕3

‌‌‌ [using Eq. (i)] ... (iii) ‌
‌ On differentiating Eq. (i) w.r.t. temperature on both sides, we get ‌
‌

dV

dT

‌‌=k⋅‌

2

3

T1∕3
‌‌=‌

2

3

kT−1∕3=‌

2

3

‌

k

T1∕3

∴‌‌dV‌‌=2∕3kT−1∕3dT
Substituting this in Eq. (iii), we get
W‌‌=∫nRT‌

2∕3kT−1∕3dT

kT2∕3

⇒‌‌W‌‌=‌

2

3

nR‌∫TdT=2∕3nR[T]T1T
‌‌=2∕3nR[T2−T1]=2∕3nR∆T
‌‌=2∕3nR90=60nR
Here, n=1
So, work done will be 60R.
Hence, x=60.