Question

when 1.4g of N2 and 4.4g of CO2 are taken in a 10L flask at 27℃ then the pressure exerted by the mixture is
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Answer 1

Given:

1.4 g of N₂ and 4.4 g of CO₂ are taken in a 10 L flask at 27℃

To find:

The pressure exerted by the mixture

Solution:

Let's assume,

"n₁" → represents the no. of moles of N₂

"n₂" → represents the no. of moles of CO₂

"P" → represents the pressure exerted by the mixture

The mass of N₂ = 1.4 g

The molar mass of N₂ = 28 g

∴ The no. of moles of N₂, n₁ = $\frac{1.4}{28} = 0.05\: mol$

The mass of CO₂ = 4.4 g

The molar mass of CO₂ = 44 g

∴ The no. of moles of CO₂, n₂ = $\frac{4.4}{44} = 0.10 \: mol$

∴ The total number of moles of the gases in the mixture is,

= n

= n₁ + n₂

= 0.05 + 0.10

= 0.15 mol

The temperature of the flask, T = 27°C = 27 + 273 K = 300 K

The volume of the flask, V = 10 L

         

We know,

The Ideal Gas Law is as follows:

$\boxed{\bold{PV = nRT}}$

where

P = pressure in atm

V = volume in litres

n = no. of moles

R = ideal gas constant = 0.0821 atm.L/mol.K

T = temperature in kelvin

Now, on substituting the values of P, V, n, R & T of the mixture of gases in the ideal gas law, we get

$PV = nRT$

$\implies P \times 10 = 0.15 \times 0.0821 \times 300$

$\implies P = \frac{0.15 \times 0.0821 \times 300}{ 10}$

$\implies \bold { P = 0.369 \: atm}$

Thus, the pressure exerted by the mixture is → 0.369 atm.

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