An ideal diatomic gas is expanded so that the amount of heat transferred to the gas is equal to the decrease in its internal energy.
If in the above process, the initial temperature of the gas be `T_(0)` and the final volume by 32 times the initial volume, the work done `(` in joules `)` by the gas during the process will be

Options

A. `RT_(0)`
B. `(5RT_(0))/(2)`
C. `2RT_(0)`
D. `(RT_(0))/(2)`

Answer

B

Explanation

Under the given process, the final volume `V` is 32 times the initial volume `V_(0)`. ltbr. Thus, `W=int_(V_(0))^(32V_(0))(PdV)/(2)=int _(V_(0)) ^(32 V_(0))(RT)/(2V)dV`
`=int_(V_(0))^(32V_(0))(R)/(2V)TdV`
Now, `TV_((1)/(5))=T_(0)V_(0)^((1)/(5))` gives` T=(T_(0)V_(0)^((1)/(5)))/(V^((1)/(5)))`
Hence,
`W=int_(V_(0))^(32V_(0))(R)/(2V)(T_(0)V_(0)^((1)/(5)))/(V^(1//5))dV`
`=RT_(0)V_(0)^(1//5)int_(V_(0))^(32V_(0))(V^(-6//5))/(2)dV`
`=RT_(0)V_(0)^(1//5)[(-5V^(-1//5))/(2)]_(V_(0))^(32V_(0))`
`W=-(5RT_(0))/(2)`
Work done by the gas `=(5rt_(0))/( 2)`