Question

As a result of isobaric heating by $\Delta T=72K$, one mole of a certain ideal gas obtains an amount of heat $Q=1.60kJ$. If the value of $\gamma$ is $\frac{(10+x)}{10}$. Find $x$.

Answer 1

By the first law of thermodynamics
$\Delta Q=\Delta U+\Delta W$
In an isobaric process
$\Delta Q=C_p\Delta T$
and $\Delta U=C_v\Delta T$ (always)

$\therefore C_p\Delta T=C_v\Delta T+\Delta W$
or $\Delta W=(C_p-C_v)\Delta T$
or $\Delta W=R\Delta T$ $(\because C_p-C_v=R)$
$\therefore \Delta W=8.3\times 72=597.6J$

$\Delta Q=C_p\Delta T$
$\therefore 1.6\times 1000=C_p\times 72$
$⟹C_p=\frac{1.6\times 1000}{72}=22.2J/molK$
$\Delta U=\Delta Q-\Delta W=1.6\times 1000-597.6=1002.4J$
But $\Delta U=C_v\Delta T$
$\therefore C_v=\frac{1002.4}{72}=13.9J/molK$
$\therefore \gamma =\frac{C_p}{C_v}=\frac{22.2}{13.9}=1.60$