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Tete edited Aug 22, 2026, 12:00 AM

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		+	### Statement
		+
		+	$5.6.28.$ [Insert the problem statement]
		+
		+	### 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,
		+
		+	\[V^{n-1}\,dT+T\cdot(n-1)V^{n-2}\,dV\qquad\Rightarrow\qquad dV=\frac{V}{T(1-n)}\,dT.\]
		+
		+	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
		+
		+	\[dQ=P\,dV+dU=\frac{5-3n}{2(1-n)}\cdot R\,dT,\]
		+
		+	so the molar heat capacity is $(5-3n)R/(2(n-1))$. It is zero when $n=5/3$ and infinite when $n=1$.
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$5.6.28.$ [Insert the problem statement]

			### 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,

			\[V^{n-1}\,dT+T\cdot(n-1)V^{n-2}\,dV\qquad\Rightarrow\qquad dV=\frac{V}{T(1-n)}\,dT.\]

			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

			\[dQ=P\,dV+dU=\frac{5-3n}{2(1-n)}\cdot R\,dT,\]

			so the molar heat capacity is $(5-3n)R/(2(n-1))$. It is zero when $n=5/3$ and infinite when $n=1$.

			#### Answer

			[Insert a concise answer or boxed result]