Question

Two moles of an ideal gas at $2$ bar and ${27}^{o}C$ expand isothermally against a constant pressure of $1$ bar. The work done by the gas is equal to:

• A. $3.5kJ$
• B. $2.5kJ$
• C. $1.5kJ$
• D. $4.5kJ$

Answer 1

Answer: (A)

$3.5kJ$

$W={\int }_{V_1}^{V^2}Pdv$

$W={\int }_{V_1}^{V_2}\left(\frac{nRT}{V}\right)dV$

$W=nRT\left[ln\left(\frac{V_2}{V_1}\right)\right]$

As we know that,

$\because$ Ideal gas and isothermal,

$\therefore P_1{V}_1=P_2{V}_2$

$\Rightarrow \frac{V_2}{V_1}=\frac{P_1}{P_2}$

$W=nRT\left[ln\left(\frac{P_1}{P_2}\right)\right]$

Given that,

$n=2$

$P_1=2bar$

$P_1=1bar$

$R=8.314J/gmolK$

$T=27℃$

$\therefore W=2\times 8.314\times 300\times \left[ln\left(\frac{2}{1}\right)\right]$

$\Rightarrow W=4988.4\times 0.693=3456.96J\approx 3.5KJ$