Work done during isothermal expansion of 1 , | vedclass

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(C) For an isothermal reversible expansion of an ideal gas,the work done $W$ is given by the formula:$W = 2.303 \, nRT \, log_{10} \left( \frac{P_1}{P

Vedclass · Accepted Answer

$1381$

Vedclass · Answer

(C) For an isothermal reversible expansion of an ideal gas,the work done $W$ is given by the formula:$W = 2.303 \, nRT \, log_{10} \left( \frac{P_1}{P_2} ight)$Given:$n = 1 \, 	ext{mole}$,$R = 2 \, 	ext{cal} \, 	ext{K}^{-1} \, 	ext{mol}^{-1}$,$T = 300 \, 	ext{K}$,$P_1 = 10 \, 	ext{atm}$,$P_2 = 1 \, 	ext{atm}$.Substituting the values:$W = 2.303 	imes 1 	imes 2 	imes 300 	imes log_{10} \left( \frac{10}{1} ight)$$W = 2.303 	imes 600 	imes 1 = 1381.8 \, 	ext{cal}$.

Work done during isothermal expansion of $1 \, \text{mole}$ of an ideal gas from $10 \, \text{atm}$ to $1 \, \text{atm}$ at $300 \, \text{K}$ is ....... $\text{cal}$ (Gas constant $R = 2 \, \text{cal} \, \text{K}^{-1} \, \text{mol}^{-1}$) (in $.8$)

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