For 2 mol of He (ideal gas),find the work do | vedclass

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(B) For an ideal gas,enthalpy $H = U + PV = nC_vT + nRT = n(C_v + R)T = nC_pT$. Since $H = 10V^2$ and $H = nC_pT$,we have $nC_pT = 10V^2$. For $He$ (m

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$-200$

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(B) For an ideal gas,enthalpy $H = U + PV = nC_vT + nRT = n(C_v + R)T = nC_pT$. Since $H = 10V^2$ and $H = nC_pT$,we have $nC_pT = 10V^2$. For $He$ (monatomic),$C_p = \frac{5}{2}R$,so $2 	imes \frac{5}{2}R 	imes T = 10V^2$,which simplifies to $5RT = 10V^2$,or $T = \frac{2}{R}V^2$. This implies $T \propto V^2$,or $TV^{-2} = 	ext{constant}$. Using the ideal gas law $PV = nRT$,we substitute $T = \frac{PV}{nR}$ into the relation: $\frac{PV}{nR}V^{-2} = 	ext{constant} \Rightarrow PV^{-1} = 	ext{constant} = K$. Work done $w = -\int_{V_1}^{V_2} P \, dV = -\int_{V_1}^{V_2} K V \, dV = -\frac{K}{2}(V_2^2 - V_1^2)$. Since $PV = nRT$,$K = PV^{-1} = \frac{nRT}{V^2}$. Substituting $V^2 = \frac{RT}{2}$ (from $T = \frac{2}{R}V^2$),we get $K = \frac{nRT}{RT/2} = 2n = 2(2) = 4$. Thus,$w = -\frac{4}{2}(V_2^2 - V_1^2) = -2(\frac{RT_2}{2} - \frac{RT_1}{2}) = -R(T_2 - T_1)$. $w = -R(400 - 200) = -200 \, R$.

For $2 \ mol$ of $He$ (ideal gas),find the work done in a process where it is heated from $200 \ K$ to $400 \ K$ such that the enthalpy of the gas varies as : $H = 10 \ V^2$ (in $R$)

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