At `27^@C` two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate (i) the final temperature of the gas, (ii) change in its internal energy, and (iii) the work done by the gas during this process.

Options

A. `179K`
B. `189K`
C. `213K`
D. `219K`

Answer

B

Explanation

Here, `T_(1)= 27+273= 300K,T_(2)=?`
`V_(1)=V` and `V_(2)= 2V, n=2`
In an adiabatic change, `T_(2)V_(2)^(gamma-1)= T_(1)V_(1)^(gamma-1)`
or `T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1)`
or `T_(2)= 300(V/(2V))^(5/3-1)= 300(0.5)^(2/3)`
On solving we get, `T_(2)= 189.03K`