A gas expands with temperature according to the relation `V = kT^(2//3)`. What is the work done when the temperature changes by `30^(@)C`?

Options

A. 10 R
B. 20 R
C. 30 R
D. 40 R

Answer

B

Explanation

`W =int P dV=int (RT)/(V) dV`
Since, `V = KT^(2//3) implies dV= 2/3 KT^(-1//3) dT`
`(dV)/(V)=2/3(dT)/(T)`
`W=int_(T_1)^(T_2) 2/3 (RT)/(T) dT=2/3 R(T_2-T_1)=2/3 R(30)=20R`