At 61 K, one mole of an ideal gas of 1.0 L | vedclass

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Question

(A) Given: $n = 1 \ mol$, $T = 61 \ K$, $V_1 = 1.0 \ L$, $V_2 = 10.0 \ L$, $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.For an isothermal reversible exp

Vedclass · Accepted Answer

$-11.52$

Vedclass · Answer

(A) Given: $n = 1 \ mol$, $T = 61 \ K$, $V_1 = 1.0 \ L$, $V_2 = 10.0 \ L$, $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.For an isothermal reversible expansion, the work done is given by the formula:$W = -nRT \ln(V_2 / V_1)$Substituting the values into the equation:$W = -(1) 	imes (0.0821) 	imes (61) 	imes \ln(10 / 1)$$W = -5.0081 	imes 2.303$$W \approx -11.53 \ L \ atm$.Rounding to the nearest provided option, the work done is $-11.52 \ L \ atm$.

At $61 \ K$, one mole of an ideal gas of $1.0 \ L$ volume expands isothermally and reversibly to a final volume of $10.0 \ L$. What is the work done in the expansion (in $L \ atm$)?

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