You visited us 1 times! Enjoying our articles? Unlock Full Access!

Graham's Law of Effusion

A gas expands...

Question

A gas expands $0.25 m^{3}$ at constant pressure $10^{3} N / m^{2}$ . The work done is

250 N

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

250 W

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

250 J

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

2.5 erg

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

Open in App

Solution

The correct option is B 250 J
Work done in expansion of gas $= \int P d V = P Δ V$
$= 10^{3} \times 0.25 = 250 J$

Suggest Corrections

Similar questions

A gas expands $0.25 m^{3}$ at constant pressure $10^{3} N / m^{2}$ , the work done is

0.25 m^{3}

10^{3} M / m^{2}

m^{3}

10^{3} N / m^{2}

10^{- 3} m^{3}

10^{- 2} m^{3}

300 K

10^{5} N / m^{2}

m^{3}

^{3} N / m^{2},

Join BYJU'S Learning Program