Calculate the work done when $2 m o l e s$ of hydrogen expand isothermally and reversibly at $25^{\circ} C$ from $15 t o 45 l i t r e s$ .

- 1309 c a l o r i e s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- 14.36 c a l o r i e s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 2872 c a l o r i e s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 28.72 c a l o r i e s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $- 1309 c a l o r i e s$
Work done in an isothermal reversible process is given as,
$w = - 2.303 n R T l o g \frac{V_{2}}{V_{1}}$
Given : $V_{1} = 15 L$
$V_{2} = 45 L$
$n = 2 m o l$
$R = 2 c a l / K . m o l$
$T = 298 K$
So,
$w = - 2.303 \times 2 \times 2 \times 298 \times l o g \frac{45}{15}$
$w = - 2745.1 \times l o g 3$
$w = - 1309.7 c a l$

Suggest Corrections

Similar questions

25^{o}

2 m o l e s

25^{\circ} C

15 t o 45 l i t r e s

25^{o} C

25^{o} C .

300 K

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program