One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure, The change in internal energy of the gas during the transition is:

Options

A. 20 kJ
B. `-20` kJ
C. 20 J
D. `-12` kJ

Answer

B

Explanation

`DeltaW=nC_V DeltaT` ….(1)
but `DeltaT=T_2-T_1`
`=(P_2V_2)/(nR)-(P_1V_1)/(nR) [ because T=(PV)/(nR)]`
`=(P_2V_2-P_1V_1)/(nR)`
Putting this value in equation (1),
`DeltaU=(nR)/(gamma-1) ((P_2V_2-P_1V_1)/(nR)) " " (because C_V=R/(gamma-1))`
`=(P_2V_2 -P_1V_1)/(gamma-1)`
`=((5xx4-2xx6)xx10^3)/(3/5-1) " " (because gamma=3/5)`
`=((20-12)xx10^3)/(-2/5)`
`=-(8xx10^3)/(2)xx5` =-20 kJ