Work done during isothermal expansion of one | vedclass

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Question

(C) The work done during the isothermal reversible expansion of an ideal gas is given by the formula:$W = 2.303 \, nRT \log_{10} \frac{P_1}{P_2}$Given

Vedclass · Accepted Answer

$5744.1$

Vedclass · Answer

(C) The work done during the isothermal reversible expansion of an ideal gas is given by the formula:$W = 2.303 \, nRT \log_{10} \frac{P_1}{P_2}$Given values:$n = 1 \, mol$,$T = 300 \, K$,$P_1 = 10 \, atm$,$P_2 = 1 \, atm$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$Substituting these values into the equation:$W = 2.303 	imes 1 	imes 8.314 	imes 300 	imes \log_{10} \left( \frac{10}{1} ight)$$W = 2.303 	imes 8.314 	imes 300 	imes 1$$W = 5744.1 \, J$Thus,the correct option is $C$.

Work done during isothermal expansion of one mole of an ideal gas from $10 \,atm$ to $1 \,atm$ at $300 \,K$ is.....$J$

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