At `27^@C` two moles of an ideal monatomic gas occupy a volume V. The gas expands adiabatically to a volume `2V`. Calculate
(a) final temperature of the gas
(b) change in its internal energy and
(c) the work done by the gas during the process. [ `R=8.31J//mol-K`]

Explanation

`n=2`,`T_1=27^@C=27+273=300K`
`V_1=V`,`V_2=2V`,`gamma=(5)/(3)`
(a) `T_1V_1^(gamma-1)=T_2V_2^(gamma-1)`
`T_2=T_1((V_1)/(V_2))^(gamma-1)=(300)((V)/(2V))^((5)/(3)-1)`
`=(300)((1)/(2))^((2)/(3))=(300)/(1.587)`
(b) `DeltaU=nC_VDeltaT=n.(3R)/(2)(T_2-T_1)`
`=2.(3)/(2)xx8.3(189-300)`
`=-2764J`
(c) `DeltaW=(nR(T_1-T_2))/(gamma-1)=(2xx8.3(300-189))/((5)/(3)-1)`
`=3xx8.3xx111=2764J`
Note: For adiabatic process, `DeltaQ=0`
`DeltaW=-DeltaU`