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

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

The mathematical equation of the first law of thermodynamics for an isochoric process is:

Calculate the internal energy change of a system if the work done by the system is $8 \ J$ and the heat supplied to it is $40 \ J$. (in $J$)

In a process,the work done by the system is equal to the change in its internal energy. Then the process will be ...........

Internal energy of an ideal gas increases in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module