Question

4 atm pressure to attain 1 atm pressure by result of isothermal expansion. The work done by the gas during expansion is nearly.

• A. 155 J
• B. 255 J
• C. 355 J
• D. 555 J

Answer 1

The correct option is D 555 J

Given,

$V=1L=0.001m^3$

$P=4atm$

$\frac{P_1}{P_2}=4$

From ideal gas,

$PV=nRT$

$T=\frac{PV}{nR}$

Work done during the isothermal expansion,

$W=nRTln\left(\frac{P_1}{P_2}\right)$

$W=nR.\frac{PV}{nR}ln\left(4\right)$

$W=PVln\left(4\right)$

$W=4\times {10}^5\times 0.001\times ln\left(4\right)$

$W=555J$

The correct option is D.