Question

As a result of the isobaric heating by $\Delta T=72K$, one mole of a certain ideal gas obtains an amount of heat Q = 1.6 kJ. Find the work performed by the gas, the increment of its internal energy and $\gamma$.

Answer 1

Given : $\Delta T=72K$ $Q=1.6$ kJ $=1600$ J

$Q=n{C}_p\Delta T$

$\therefore$ $1600=1\times C_p\times 72$ $⟹C_p=22.22$ $J/K-mol$

Also from $C_P-C_V=R$

$\therefore$ $22.22-C_V=8.314$ $⟹C_V=13.91$ $J/K-mol$

Thus $\gamma =\frac{C_P}{C_V}=\frac{22.22}{13.91}=1.6$

Change in internal energy $\Delta U=n{C}_V\Delta T=1\times 13.91\times 72=1000$ J

Using $Q=\Delta U+W$

$\therefore$ $1600=1000+W$ $⟹W=600$ J