The pressure of $2$ moles of ideal gas at $546 K$ having volume $44.8 L$ is:

2

atm

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

3

atm

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

4

atm

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

1

atm

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

Open in App

Solution

The correct option is A $2$ atm
$P V = n R T P \frac{n R T}{V} = \frac{2 \times 0.08 \times 546}{44.8} = 2 a t m$ Given $n = 2, T = 546 K, V = 44.8 L$

Suggest Corrections

Similar questions

Q. If

2

moles of an ideal gas at

546 K

occupy volume

44.8 L

, then pressure must be :

Q. 1/2 mole oxygen gas occupies 44.8 L at temperature 273.15 K. The pressure of the oxygen gas is .

Q. Oxygen gas which is at a pressure of

2^{3 / 5} atm

is compressed adiabatically so that volume becomes half that of the initial volume. The final pressure of the gas is

The equilibrium constant Kp of the reaction, 2 $S O_{2}$ (g) + $O_{2}$ (g) $\leftrightharpoons$ 2 $S O_{3}$ (g) is 900 atm. at 800 K. A mixture containing $S O_{3}$ and $O_{2}$ having initial pressure of 1 and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of $S O_{2}$ gas at 800 K.

5 moles

of an ideal gas expand isothermally and irreversibly from a pressure of

10 atm

1 atm

against a constant external pressure of

1 atm

W_{i r r}

300 K

is:

Join BYJU'S Learning Program

The pressure of 2 moles of ideal gas at 546K having volume 44.8L is:

The correct option is A 2 atmPV=nRTPnRTV=2×0.08×54644.8=2atm Given n=2,T=546K,V=44.8L

The pressure of $2$ moles of ideal gas at $546 K$ having volume $44.8 L$ is:

The correct option is A $2$ atm
$P V = n R T P \frac{n R T}{V} = \frac{2 \times 0.08 \times 546}{44.8} = 2 a t m$ Given $n = 2, T = 546 K, V = 44.8 L$