| @@ -1,23 +1,23 @@ |
| ### Statement |
| ### Statement |
| |
| |
| $5.6.28.$ [Insert the problem statement] |
| $5.6.28.$ Find the molar heat capacity of a monatomic gas expanding according to the law $PV^n=$ const. At what values of will the heat capacity be zero? infinity? |
| |
| |
| ### Solution |
| ### Solution |
| |
| |
| Since $PV^n$ is constant and so is $PV/T$ for ideal gas, it follows that $TV^{n-1}$ is constant. As a result, |
| Since $PV^n$ is constant and so is $PV/T$ for ideal gas, it follows that $TV^{n-1}$ is constant. As a result, |
| |
| |
| \[V^{n-1}\,dT+T\cdot(n-1)V^{n-2}\,dV\qquad\Rightarrow\qquad dV=\frac{V}{T(1-n)}\,dT.\] |
| \[V^{n-1}\,dT+T\cdot(n-1)V^{n-2}\,dV=0\qquad\Rightarrow\qquad dV=\frac{V}{T(1-n)}\,dT.\] |
| |
| |
| Consequently, the work done by one mole of gas in an expansion over a small temperature change is |
| Consequently, the work done by one mole of gas in an expansion over a small temperature change is |
| |
| |
| \[P\,dV=\frac{PV}{T(1-n)}\,dT=\frac{R}{1-n}\,dT.\] |
| \[P\,dV=\frac{PV}{T(1-n)}\,dT=\frac{R}{1-n}\,dT.\] |
| |
| |
| At the same time, the change in internal energy of monatomic gas is $dU=(3R/2)\,dT$. Thus, the gas must absorb heat |
| At the same time, the change in internal energy of monatomic gas is $dU=(3R/2)\,dT$. Thus, the gas must absorb heat |
| |
| |
| \[dQ=P\,dV+dU=\frac{5-3n}{2(1-n)}\cdot R\,dT,\] |
| \[dQ=P\,dV+dU=\frac{(5-3n)R}{2(1-n)}\,dT,\] |
| |
| |
| so the molar heat capacity is $(5-3n)R/(2(n-1))$. It is zero when $n=5/3$ and infinite when $n=1$. |
| so the molar heat capacity is $(5-3n)R/(2(1-n))$. It is zero when $n=5/3$ and infinite when $n=1$. |
| |
| |
| #### Answer |
| #### Answer |
| |
| |
| [Insert a concise answer or boxed result] |
| $(5-3n)R/(2(1-n))$, which is zero when $n=5/3$ and infinite when $n=1$. |
