Question
When one mole of monoatomic ideal gas at $$T\,K$$ undergoes adiabatic change under a constant external pressure of 1 atm volume changes from 1 litre to 2 litre. The final temperature in Kelvin would be
A.
$$\frac{T}{{{2^{\left( {\frac{2}{3}} \right)}}}}$$
B.
$$T + \frac{2}{3} \times 0.0821$$
C.
$$T$$
D.
$$T - \frac{2}{3} \times 0.0821$$
Answer :
$$\frac{T}{{{2^{\left( {\frac{2}{3}} \right)}}}}$$
Solution :
$$\eqalign{
& T{V^{\gamma - 1}} = {\text{Constant}}\,\,\,\,\,\,\,\left( {\because \,\,{\text{Change}}\,{\text{is}}\,{\text{adiabatic}}} \right) \cr
& {T_1}{V_1}^{^{\gamma - 1}} = {T_2}V_2^{^{\gamma - 1}} \cr
& {\text{For monoatomic gas}}\,\,\gamma = \frac{5}{3} \cr
& \therefore \,\,{T_1}V_1^{\frac{2}{3}} = {T_2}V_2^{\frac{2}{3}} \Rightarrow T{\left( 1 \right)^{\frac{2}{3}}} = {T_2}{\left( 2 \right)^{\frac{2}{3}}} \cr
& {T_2} = \frac{T}{{{2^{\left( {\frac{2}{3}} \right)}}}} \cr} $$