One mole of monoatomic ideal gas at `T(K)` is exapanded from `1L` to `2L` adiabatically under constant external pressure of `1` atm . The final tempreture of gas in kelvin is

Options

A. `T`
B. `(T)/(2^(5//3-1))`
C. `T-(2)/(3xx0.0821)`
D. `T+(2)/(3xx0.0821)`

Answer

C

Explanation

Here `P_("ext")=1"atm"`
`V_(1)=1L ` and `V_(2)=2L`
`R=0.0821 L "atm" K^(-1) mol^(-1)`
`gamma=5//3` (For a monoatomic gas)
`T_(1)=T K` and `T_(2) = ?`
Now `C_(P)-C_(V)=R` or `(C_(P))/(C_(V))-1=(R )/(C_(V))`
`gamma-1=(R )/(C_(V))`
`C_(V)=(R )/(gamma-1)`
`C_(V)(T_(2)-T_(1))=P_("ext")(V_(1)-V_(2))`
`(R )/(gamma-1)(T_(2)-T_(1))=P_("ext")(V_(1)-V_(2))`
Substiuting the values
`(0.0821)/(5//3-1)(T_(2)-T_(1))=1xx(!-2)`
`T_(2)-T=(2//3)/(0.0821)`
`T_(2)=T-(2)/(3xx0.0821)`