Новое решение

JMMA2006 правка от 19 июн. 2026 г., 17:06

правка #19037 позже →

en/5.6.17.md +30 −0

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		+	### Statement
		+
		+	$5.6.17.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	Assuming that hydrogen (H_2) behaves has an ideal gas\
		+	We have\
		+	$n=1mol$\
		+	$T=0\degree C=273K$\
		+	$P=const$\
		+	Q=nC_p\Delta T\
		+	we know that $C_p=\frac{(i+2)R}{2}$\
		+	where i is the number of degrees of freedom, which is 5 for hydrogen (diatomic gas)\
		+	$C_p=\frac{7R}{2}$\
		+	so $Q=\frac{7nR\Delta T}{2}$\
		+	and with Charles's Law ( $\frac{V}{T}=const$ )\
		+	$\frac{V}{T_1}=\frac{2V}{T_2}$\
		+	$T_2=2T_1$ , so $\Delta T=T_2-T_1=2T_1-T_1=T_1$\
		+	and substituting in the heat\
		+	$Q=\frac{7nRT_1}{2}$\
		+	$Q\approx 7948.5J$
		+	and the work for a constant pressure\
		+	$W=P\Delta V=P(2V-V)=PV=nRT_1$\
		+	$W\approx 2271J$\
		+	It's an isobaric expansion work
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,30 @@
			### Statement

			$5.6.17.$ [Insert the problem statement]

			### Solution

			Assuming that hydrogen (H_2) behaves has an ideal gas\
			We have\
			$n=1mol$\
			$T=0\degree C=273K$\
			$P=const$\
			Q=nC_p\Delta T\
			we know that $C_p=\frac{(i+2)R}{2}$\
			where i is the number of degrees of freedom, which is 5 for hydrogen (diatomic gas)\
			$C_p=\frac{7R}{2}$\
			so $Q=\frac{7nR\Delta T}{2}$\
			and with Charles's Law ( $\frac{V}{T}=const$ )\
			$\frac{V}{T_1}=\frac{2V}{T_2}$\
			$T_2=2T_1$ , so $\Delta T=T_2-T_1=2T_1-T_1=T_1$\
			and substituting in the heat\
			$Q=\frac{7nRT_1}{2}$\
			$Q\approx 7948.5J$
			and the work for a constant pressure\
			$W=P\Delta V=P(2V-V)=PV=nRT_1$\
			$W\approx 2271J$\
			It's an isobaric expansion work

			#### Answer

			[Insert a concise answer or boxed result]