One mole of an ideal gas at 300 K and 20 at | vedclass

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Question

(A) For an isothermal reversible expansion of an ideal gas, the work done $(w)$ is given by the formula: $w = -nRT \ln(\frac{P_1}{P_2})$.Given: $n = 1

Vedclass · Accepted Answer

$5.73$

Vedclass · Answer

(A) For an isothermal reversible expansion of an ideal gas, the work done $(w)$ is given by the formula: $w = -nRT \ln(\frac{P_1}{P_2})$.Given: $n = 1 \ mol$, $T = 300 \ K$, $P_1 = 20 \ atm$, $P_2 = 2 \ atm$, and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$.Substituting the values: $w = -1 	imes 8.3 	imes 300 	imes \ln(\frac{20}{2})$.$w = -2490 	imes \ln(10)$.Using $\ln(10) \approx 2.303$: $w = -2490 	imes 2.303 = -5734.47 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$: $w = -5.734 \ kJ \ mol^{-1}$.Since the work done is $-x \ kJ \ mol^{-1}$, we have $-x = -5.734$, so $x = 5.73$.

One mole of an ideal gas at $300 \ K$ and $20 \ atm$ expands to $2 \ atm$ under isothermal and reversible conditions. The work done by the gas is $-x \ kJ \ mol^{-1}$. The value of $x$ is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

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