Edit to “Statement”
en/5.6.28.md
+1 −1
| @@ -1,6 +1,6 @@ | |||
| ### 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? | ||
| + | $5.6.28^*.$ 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 | |||
| Since $PV^n$ is constant and so is $PV/T$ for ideal gas, it follows that $TV^{n-1}=const$. Differentiating both sides using the product rule $d(uv) = v\,du + u\,dv$, we obtain: | |||
| \[V^{n-1}\,dT+T\cdot(n-1)V^{n-2}\,dV=0\qquad\Rightarrow\qquad dV=\frac{V}{T(1-n)}\,dT.\] | |||
| Consequently, the work done by one mole of gas in an expansion over a small temperature change is | |||
| \[P\,dV=\frac{PV}{T(1-n)}\,dT=\frac{R}{1-n}\,dT.\] | |||
| At the same time, the change in internal energy of monatomic gas is $dU=(3R/2)\,dT$. Thus, the gas must absorb heat | |||
| \[dQ=P\,dV+dU=\frac{(5-3n)R}{2(1-n)}\,dT,\] | |||
| 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 | |||
| $(5-3n)R/(2(1-n))$, which is zero when $n=5/3$ and infinite when $n=1$. | |||
| unchanged lines 17 | |||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### 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? | $5.6.28^*.$ 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 $TV^{n-1}=const$. Differentiating both sides using the product rule $d(uv) = v\,du + u\,dv$, we obtain: | Since $PV^n$ is constant and so is $PV/T$ for ideal gas, it follows that $TV^{n-1}=const$. Differentiating both sides using the product rule $d(uv) = v\,du + u\,dv$, we obtain: | ||
| \[V^{n-1}\,dT+T\cdot(n-1)V^{n-2}\,dV=0\qquad\Rightarrow\qquad dV=\frac{V}{T(1-n)}\,dT.\] | \[V^{n-1}\,dT+T\cdot(n-1)V^{n-2}\,dV=0\qquad\Rightarrow\qquad dV=\frac{V}{T(1-n)}\,dT.\] | ||
| 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 | ||
| \[dQ=P\,dV+dU=\frac{(5-3n)R}{2(1-n)}\,dT,\] | \[dQ=P\,dV+dU=\frac{(5-3n)R}{2(1-n)}\,dT,\] | ||
| so the molar heat capacity is $(5-3n)R/(2(1-n))$. 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 | ||
| $(5-3n)R/(2(1-n))$, which is zero when $n=5/3$ and infinite when $n=1$. | $(5-3n)R/(2(1-n))$, which is zero when $n=5/3$ and infinite when $n=1$. | ||
| unchanged lines 17 | |||