2 moles of helium gas expanded isothermally and irreversible at $27$ from volume $1 {d m}^{3}$ to $1 m^{3}$ at constant pressure of $100 k P a$ . Calculate the work done.

99900 k J

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

99900 J

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

34464.65 k J

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

34464.65 J

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

Open in App

Solution

The correct option is B $99900 J$
$W = - p Δ V$
Given, $p = 100 k P a = 10^{5} P a,$
$V_{1} = 1 {d m}^{3} = 10^{- 3} m^{3}, V_{2} = 1 m^{3}$
$W = 10^{5} \times (1 - 10^{- 3}) J$
$∴ W = 99900 J$

Suggest Corrections

Similar questions

27^{0} C

1 d m^{3}

1 m^{3}

100 K P a .

The work done when two moles of an ideal gas is compressed from a volume of 5 m³ to 1 dm³ at 300K,under a pressure of 100kPa?

300 c m^{3}

2.5 L

300 K

1.9

300 K

6

1 {d m}^{3}

10 {d m}^{3}

27^{o} C

Join BYJU'S Learning Program