Discussion Forum

Two moles of an ideal monoatomic gas occupies a volume V at 27° C. The gas expands adiabatically to a Volume 2 V. Calculate(i) the final temperature of the gas and (ii) change in its internal energy.

Answer:

Explanation:

For adiabatic process relation of temperature and volume is,

$T_{2}V_{2}^{\gamma-1}=T_{1}V_{1}^{\gamma-1}$

$T_{2}(2V_{})^{\frac{2}{3}}=300_{}(V_{})^{\frac{2}{3}}$

           [ $\gamma=\frac{5}{3} $ for monoatomic gases ]

$T_{2}=\frac{300}{2^{\frac{2}{3}}}\approx 189K$

 Also, in adiabatic process,

  $\triangle Q=0, \triangle U=-\triangle W$

 OR       $\triangle U=\frac{-nR(\triangle T)}{\gamma -1}$

  $= -2\times \frac{3}{2}\times \frac{25}{3}(300-189)$

    $\approx -2.7KJ$

     $T_{2}\approx 189K,\triangle U\approx2.7KJ$