Question

One mole of an ideal monoatomic gas has internal energy 18000 J. The gas was cooled isochorically till the gas pressure fell from p to p/3. Then, by an isobaric process, the gas was restored to the initial temperature. Find the net amount of heat absorbed (in KJ) by the gas in the process.

Answer 1

During an isochoric process,

$\Delta W=0$

$\Delta Q=\Delta W=0$

$\therefore$ Heat absorbed in isobaric process

$P=\frac{P}{3}$

$Q=\Delta U+\Delta W$

$=\Delta U+P\Delta V$

$P\Delta V=nR\Delta T$

$\Delta V=\frac{n}{P}R\Delta T\therefore Q=\Delta U+nR\Delta T=18000+\left(1\right)\left(8.31\right)\Delta T=[18000+8.31\Delta T]kJ$