Work done during isothermal expansion of one | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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.....$J$

Similar Questions

In which of the following reactions is $\Delta E = \Delta H$?

Which of the following is an intensive property?

Which expression is correct for the work done in adiabatic reversible expansion of an ideal gas?

The internal energy of a substance

How will you calculate the work done on an ideal gas during compression,when the change in pressure is carried out in infinite steps?

Vedclass Test Series

Exam Paper Generator

Online Exam Module