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.