$5 moles$ of an ideal gas expand isothermally and irreversibly from a pressure of $10 atm$ to $1 atm$ against a constant external pressure of $1 atm$ . $W_{i r r}$ at $300 K$ is:

- 15.921 kJ

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

- 11.224 kJ

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

- 110.83 kJ

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

None of these

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

Open in App

Solution

The correct option is B $- 11.224 kJ$
$\int d W = - \int P_{e x t} . d V ∴ W_{i r r} = - P_{e x t} [\frac{n R T}{P_{2}} - \frac{n R T}{P_{1}}] \Rightarrow W_{i r r} = - 1 \times (5 \times 8.314 \times 300) \times [1 - \frac{1}{10}] \Rightarrow W_{i r r} = - 11.224 kJ$

Suggest Corrections

Similar questions

5 mole of an ideal gas expand isotherally and irreversibly from a pressure of 10 atm to 1 atm against a contant external pressure of 1 at. Work done at 300 K is:

Q. Two mole of ideal diatomic gas

(C_{v, m} = 5 / 2 R)

300 K

and

10 a t m

expanded irreversibly and adiabatically to a final pressure of

2 a t m

against a constant pressure of 1 atm. Calculate change in internal energy

△ U

Q. Two mole of ideal diatomc gas (

C_{v, m} = \frac{5}{2} R

) at 300 K and 5 atm is expanded irreversibly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. Calculate q, w, H and U.

Q. Four moles of an ideal gas at

27^{\circ} C

is allowed to expand isothermally from an initial pressure of

12.0 atm

to a final pressure of

6.0 atm

against a constant external pressure of

1.0 atm

. Thus, work done will be

$2.8 g$ of $N_{2}$ gas at $300$ K and $20$ atm was allowed to expand isothermally against a constant external pressure of $1$ atm. Calculate $W$ for the gas.

Join BYJU'S Learning Program

5 moles of an ideal gas expand isothermally and irreversibly from a pressure of 10 atm to 1 atm against a constant external pressure of 1 atm. Wirr at 300 K is:

The correct option is B −11.224 kJ∫dW=−∫Pext.dV∴Wirr=−Pext[nRTP2−nRTP1]⇒Wirr=−1×(5×8.314×300)×[1−110]⇒Wirr=−11.224 kJ

$5 moles$ of an ideal gas expand isothermally and irreversibly from a pressure of $10 atm$ to $1 atm$ against a constant external pressure of $1 atm$ . $W_{i r r}$ at $300 K$ is:

The correct option is B $- 11.224 kJ$
$\int d W = - \int P_{e x t} . d V ∴ W_{i r r} = - P_{e x t} [\frac{n R T}{P_{2}} - \frac{n R T}{P_{1}}] \Rightarrow W_{i r r} = - 1 \times (5 \times 8.314 \times 300) \times [1 - \frac{1}{10}] \Rightarrow W_{i r r} = - 11.224 kJ$