the volume of a gas expands by 0.25 $m^{3}$ at a constant pressure of $10^{3} N / m^{2}$ . The work done is equal to

2.5 erg

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

250 w

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

250 N

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

Open in App

Solution

The correct option is B 250 j
Given,
$d V = 0.25 m^{3}$
$P = 10^{3} N / m^{2}$
Work done, $W = \int P d V$
$W = 0.25 \times 1000 = 250 J$
The correct option is B.

Suggest Corrections

Similar questions

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

m^{3}

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

0.25 m^{3}

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

10^{- 3} m^{3}

10^{- 2} m^{3}

300 K

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

1 \times 10^{- 3} m^{3}

1 \times 10^{- 2} m^{3}

300 K

1 \times 10^{5} {N m}^{- 2}

Join BYJU'S Learning Program