One mole of an ideal gas expands isothermally | vedclass

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(B) For an isothermal process,$\Delta T = 0$,therefore the change in internal energy $\Delta U = 0$. For a reversible isothermal expansion,the work do

Vedclass · Accepted Answer

$0, 1.718 \ kJ, -1.718 \ kJ$

Vedclass · Answer

(B) For an isothermal process,$\Delta T = 0$,therefore the change in internal energy $\Delta U = 0$. For a reversible isothermal expansion,the work done $w = -nRT \ln(V_2/V_1)$. Given $n = 1 \ mol$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_1 = 10 \ dm^3$,$V_2 = 20 \ dm^3$. $w = -1 	imes 8.3 	imes 300 	imes \ln(20/10) = -2490 	imes \ln(2)$. Using $\ln(2) = 2.303 	imes \log(2) = 2.303 	imes 0.30 = 0.6909$. $w = -2490 	imes 0.6909 = -1720.34 \ J \approx -1.718 \ kJ$. Since $\Delta U = q + w = 0$,we have $q = -w = 1.718 \ kJ$. Thus,$\Delta U = 0, q = 1.718 \ kJ, w = -1.718 \ kJ$.

One mole of an ideal gas expands isothermally and reversibly from $10 \ dm^3$ to $20 \ dm^3$ at $300 \ K$. $\Delta U$,$q$ and work done in the process respectively are $:$ Given $: R=8.3 \ J \ K^{-1} \ mol^{-1}$,$\ln 10=2.3$,$\log 2=0.30$,$\log 3=0.48$

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