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A gas expands with temperature according...

A gas expands with temperature according to the relation `V=KT^(2/3)`.Work done when the temperature changes by 60K is.

A

10 R

B

20 R

C

30 R

D

40 R

Text Solution

Verified by Experts

The correct Answer is:
B
`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`
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