You visited us 21 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Real Gas Equation

Total pressur...

Question

Total pressure of a gaseous mixture containing $28 g N_{2}$ and $16 g O_{2}$ in a cylinder of volume $8.21 L$ at $300 K$ is:
[Given: $R = 0.0821 {L atm mol}^{- 1} K^{- 1}$ ]

3.5 atm

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

2.5 atm

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

4.5 atm

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

5.5 atm

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

Open in App

Solution

The correct option is C $4.5 atm$
$P = \frac{n R T}{V} = \frac{(\frac{28}{28} + \frac{16}{32}) \times 0.0821 \times 300}{8.21} = 4.5 atm$

Suggest Corrections

Similar questions

5 g

of a gaseous mixture of the

H e

and

A r

contained in a

10 L

vessel at

27 ° C

exert a pressure of

1 atm

(absolute). Determine the mass percentage of

H e

.
(Take

R = \frac{1}{12} {L atm mol}^{- 1} K^{- 1}

)

Q. A mixture of

N_{2}

and

O_{2}

gases in a cylinder contains

7 g

N_{2}

and

8 g

O_{2}

. If the total pressure of the mixture of the gases in the cylinder is

12 bar

, then the partial pressure of

N_{2}

is:
[Use atomic masses,

N = 14 u, O = 16 u

]

Q. A mixture of

{N H}_{3} (g)

and

N_{2} H_{4} (g)

is placed in a sealed container at 300 K .The total pressure of the gaseous mixture is 0.5atm . The container is heated to 1200 K at which both

{N H}_{3}

and

N_{2} H_{4}

decomposed completely according to the following equation.

2 {N H}_{3} (g) ⟶ N_{2} (g) + 3 H_{2} (g)

N_{2} H_{4} (g) ⟶ N_{2} (g) + 2 H_{2} (g)

After decomposition is complete, the total pressure at 1200 K is found to be 4.5 atm, The mole % of

N_{2} H_{4}

in the initial mixture at 300 K was:

A gaseous mixture containing 0.35 g of N2 and 5600ml of O2 at STP is kept in a 5litre flask at 300K. The total pressure of the gaseous mixture is?

Q. Find out the volume of solution containing

2 mol

of sugar at an osmotic pressure of

1 a t m and temperature a t 27^{\circ} C

?
(Given,

R = 0.0821 {L atm mol}^{- 1} K^{- 1}

)

Join BYJU'S Learning Program

Total pressure of a gaseous mixture containing 28 g N2 and 16 g O2 in a cylinder of volume 8.21 L at 300 K is: [Given: R=0.0821 L atm mol−1K−1]

The correct option is C 4.5 atmP=nRTV=(2828+1632)×0.0821×3008.21=4.5 atm

Total pressure of a gaseous mixture containing $28 g N_{2}$ and $16 g O_{2}$ in a cylinder of volume $8.21 L$ at $300 K$ is:
[Given: $R = 0.0821 {L atm mol}^{- 1} K^{- 1}$ ]

The correct option is C $4.5 atm$
$P = \frac{n R T}{V} = \frac{(\frac{28}{28} + \frac{16}{32}) \times 0.0821 \times 300}{8.21} = 4.5 atm$