An ideal gas at $${27^ \circ }C$$  is compressed adiabatically to $$\frac{8}{{27}}$$ of its original volume. The rise in temperature is $$\left( {\gamma = \frac{5}{3}} \right)$$

Solution :
In an adiabatic process
$$p =$$  pressure
$$V =$$  volume
$$\gamma = $$  atomicity of gas
$$p{V^\gamma } = {\text{constant}}\,......\left( {\text{i}} \right)$$
Now from ideal gas equation,
$$\eqalign{ & pV = RT\,\,\,\,\left( {{\text{for}}\,{\text{one}}\,{\text{mole}}} \right) \cr & {\text{or}}\,\,p = \frac{{RT}}{V}\,......\left( {{\text{ii}}} \right)\,\left( {R = {\text{gas}}\,{\text{constant}}} \right) \cr} $$
From Eqs. (i) and (ii), we have
$$\eqalign{ & \left( {\frac{{RT}}{V}} \right){V^\gamma } = {\text{constant}} \cr & T{V^{\gamma - 1}} = {\text{constant}}\,......\left( {{\text{iii}}} \right) \cr} $$
So for two different cases of temperature and volume
$$\eqalign{ & {\text{So,}}\,\,{T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1} \cr & {\text{or}}\,\,\frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}\,......\left( {{\text{iv}}} \right) \cr & {\text{Given,}}\,\,{T_1} = {27^ \circ }C = 27 + 273 = 300\,K \cr & {\text{Given,}}\,\,\frac{{{V_2}}}{{{V_1}}} = \frac{8}{{27}},\gamma = \frac{5}{3} \cr} $$
Substituting in Eq. (i), we get
$$\eqalign{ & \frac{{{T_2}}}{{300}} = {\left( {\frac{{27}}{8}} \right)^{\frac{5}{3} - 1}} \cr & {\text{or}}\,\,\frac{{{T_2}}}{{300}} = {\left[ {{{\left( {\frac{3}{2}} \right)}^3}} \right]^{\frac{2}{3}}}{\text{ or }}\frac{{{T_2}}}{{300}} = {\left( {\frac{3}{2}} \right)^2} = \frac{9}{4} \cr & \therefore {T_2} = \frac{9}{4} \times 300 = 675\;K = {402^ \circ }C \cr} $$
Thus, rise in temperature $$ = {T_2} - {T_1} = 402 - 27 = {375^ \circ }C$$