The work done in ergs for the reversible expa | vedclass

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

Question

(B) For a reversible isothermal expansion of an ideal gas,the work done is given by the formula: $W = -2.303 nRT \log(V_2/V_1)$.Here,$n = 1 \ mol$,$T

Vedclass · Accepted Answer

$-298 	imes 10^7 	imes 8.31 	imes 2.303 	imes \log 2$

Vedclass · Answer

(B) For a reversible isothermal expansion of an ideal gas,the work done is given by the formula: $W = -2.303 nRT \log(V_2/V_1)$.Here,$n = 1 \ mol$,$T = 25^{\circ} C = 298 \ K$,$V_1 = 10 \ L$,and $V_2 = 20 \ L$.The gas constant $R$ in $erg \ K^{-1} \ mol^{-1}$ is $8.314 	imes 10^7 \ erg \ K^{-1} \ mol^{-1}$.Substituting the values: $W = -2.303 	imes 1 	imes (8.314 	imes 10^7) 	imes 298 	imes \log(20/10)$.$W = -2.303 	imes 8.314 	imes 10^7 	imes 298 	imes \log 2$.This matches the expression in option $B$.

The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of $10 \ L$ to $20 \ L$ at $25^{\circ} C$ is :

Similar Questions

$A$ certain mass of an ideal gas absorbs $80 \ kJ$ heat and the gas is expanded from $2 \ L$ to $10 \ L$ at a constant pressure of $25 \ bar$. What is $\Delta U$ for the gas in the process in $kJ$? $(1 \ bar-L = 100 \ J)$

The reaction in which $\Delta H > \Delta U$ is

$A$ mixture of $2 \ mol$ of carbon monoxide and $1 \ mol$ of oxygen is ignited in a closed vessel to form carbon dioxide. If $\Delta H$ is the enthalpy change and $\Delta U$ is the change in internal energy,then:

Which of the following statements is appropriate as per the first law of thermodynamics?

$A$ gas occupies $1.5 \ L$ at $S.T.P.$ It is provided with $300 \ J$ of heat so that its volume becomes $2 \ L$ at a pressure of $1 \ atm$. The value of $\Delta U$ for the process is.....joule.

Vedclass Test Series

Exam Paper Generator

Online Exam Module