A gas expands with temperature according to the relation, `V=kT^(2//3)` where k is a constant Word done when the temperature changes by 60 K is (R= Universal gas constant).

Options

A. 10 R
B. 20 R
C. 50 R
D. 40 R

Explanation

Given, `V=kT^(2//3)`
From definition of word done,
`dW=P d V=(RT)/V dV`
`=(RT)/(kT^(2//3)) dV.....(i)`
Now `V=kT^(2//3)`
Taking derivative on the both sides, we get
`dV=K2/3T^(1//3)dT`.......(ii)
Subsituting the value of Eqs. (ii) in Eqs. (i) and taking integration on the both sides, we get
`W=2/3R int_(T_1)^(T_2) dT=2/3R(T_2-T_1)`
`=2/3 R(60-0)=2/3 timesR times 60=40 R`