Question

1 mole of gas expands isothermally at ${37}^{\circ }$ C. the amount of heat is absorbed by it until its volume doubled is ( R = 8.31 J ${mol}^{-1}{K}^{-1}$ )

• A. 411.25 cal
• B. 418.50 cal
• C. 420.25 cal
• D. 425.40 cal

Answer 1

The correct option is D 425.40 cal

Given,

$n=1$

$T={37}^{0}C=273+37=310K$

$V_1=V$

$V_2=2V$

$R=8.31J/molK$

In isothermal process,

$W=nRTln\left(\frac{V_2}{V_1}\right)$

$W=1\times 8.31\times 310\times ln\left(\frac{2V}{V}\right)$

$W=8.31\times 310\times ln\left(2\right)$

$W=2576.1\times 0.693$

$W=1785.6J$

We know that, $1cal=4.2J$

$1J=\frac{1}{4.2}cal$

$1785.6J=\frac{1785.6}{4.2}=425.4cal$

The amount of heat is $425.4cal$

The correct option is D.