The work done during reversible isothermal ex | vedclass

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

Question

(A) For a reversible isothermal expansion,the work done is given by the formula:$W = -2.303 \ nRT \log \frac{P_i}{P_f}$Given:$n = 1 \ mol$$R = 2.0 \ c

Vedclass · Accepted Answer

$-413.14 \ calories$

Vedclass · Answer

(A) For a reversible isothermal expansion,the work done is given by the formula:$W = -2.303 \ nRT \log \frac{P_i}{P_f}$Given:$n = 1 \ mol$$R = 2.0 \ cal \ K^{-1} \ mol^{-1}$$T = 25^{\circ}C = 298 \ K$$P_i = 20 \ atm$$P_f = 10 \ atm$Substituting the values:$W = -2.303 	imes 1 	imes 2.0 	imes 298 	imes \log \frac{20}{10}$$W = -2.303 	imes 2.0 	imes 298 	imes \log 2$$W = -2.303 	imes 596 	imes 0.3010$$W \approx -413.14 \ calories$

The work done during reversible isothermal expansion of one mole of hydrogen gas at $25^{\circ}C$ from a pressure of $20 \ atm$ to $10 \ atm$ is
(Given $R = 2.0 \ cal \ K^{-1} \ mol^{-1}$)

Similar Questions

At $27\,^{\circ}C$,$1\, mole$ of an ideal gas expands reversibly and isothermally from $2\, atm$ to $1\, atm$. Then $\Delta H$ will be......$kJ$.

Two moles of an ideal gas are expanded isothermally from $15 \ dm^3$ to $20 \ dm^3$. If the amount of work done is $-6 \ dm^3 \ bar$,find the external pressure needed to obtain this work.

For the isothermal expansion of an ideal gas:

For which of the following reactions,$\Delta H$ is equal to $\Delta U$ ?

Which one of the following reactions has $\Delta H = \Delta U$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The work done during reversible isothermal expansion of one mole of hydrogen gas at $25^{\circ}C$ from a pressure of $20 \ atm$ to $10 \ atm$ is(Given $R = 2.0 \ cal \ K^{-1} \ mol^{-1}$)