# Heat Capacity of an ideal gas [duplicate]

This question already has an answer here:

The question is:

A monoatomic gas undergoes a process $$PV^3=k$$ where k is a constant. Find the heat capacity of the gas.

Problem:

I know how to calculate the heat capacity at constant pressure ($$C_p$$) and ($$C_v$$) of a gas but this problem does not ask any of them.

For a process $$PV^n=k$$, $$C = (\frac{1}{\gamma - 1}-\frac{1}{n-1})R$$

as the solution but found no proof or anything. How can I start to prove it myself?

## marked as duplicate by sammy gerbil, harshit54, Jon Custer, John Rennie thermodynamics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 6 '18 at 4:34

If $$PV^n=k$$, then $$P=\frac{P_0V_0^n}{V^n}$$and $$PdV=\frac{P_0V_0^n}{V^n}dV=-\frac{P_0V_0^n}{(n-1)}d\left(\frac{1}{V^{n-1}}\right)$$So, $$\int_{V_0}^V{PdV}=\frac{P_0V_0^n}{(n-1)}\left[\frac{1}{V_0^{n-1}}-\frac{1}{V^{n-1}}\right]=\frac{P_0V_0^n}{(n-1)}\left[\frac{V_0}{V_0^{n}}-\frac{V}{V^{n}}\right]=\frac{P_0V_0-PV}{(n-1)}$$From this, it follows that, for this particular process path $$PdV=-\frac{d(PV)}{(n-1)}=-\frac{RdT}{(n-1)}$$So, from the first law: $$dU=C_vdT=dQ+\frac{RdT}{(n-1)}$$So, $$dQ=\left[C_v-\frac{R}{(n-1)}\right]dT$$ The authors refer to the term in brackets as the "heat capacity," but it should not really be considered heat capacity, since, in thermodynamics, we regard heat capacity is a fundamental physical property of the gas (independent of process path).
• I think it makes sense because we also define $C_p$ and $C_v$ in a similar fashion. So calling this heat capacity for this process makes sense. – harshit54 Oct 5 '18 at 17:58