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Thermodynamics Part 3

Section: Physics

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Question : 74 of 100

Marks: +1, -0

A cylindrical container of volume

4.0 \times 10^{-3} \text{m}^3

contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is

400 \text{K}

. The pressure of the mixture of gases is

[Take, gas constant

=8.3 \text{J} \text{mol}^{-1} \text{K}^{-1}

[26 Aug 2021 Shift 2]

$249 \times 10^{1} \text{Pa}$
$24.9 \times 10^{3} \text{Pa}$
$24.9 \times 10^{5} \text{Pa}$
$24.9 \text{Pa}$

Solution:

Given, volume of container,

V=4.0 \times 10^{-3} \text{m}^3

Gas constant,

R=8.3 \text{J} \; \text{mol}^{-1} \text{K}^{-1}

Number of moles of hydrogen,

n_1=1

Number of moles of carbon dioxide,

n_2=2

Temperature of mixture, T = 400 K

Using Dalton’s law to calculate the pressure of mixture as follows

p=p_1+p_2

=\frac{n_1 R T}{V}+\frac{n_2 R T}{V}

=\frac{R T}{V}(n_1+n_2)

= \frac{R T}{V}(1+2)=\frac{3 R T}{V}

=\frac{3 \times 8.3 \times 400}{4.0 \times 10^{-3}}=24.9 \times 10^{5} \text{Pa}

Thus, the pressure of mixture of gases is

24.9 \times 10^{5} \text{Pa}

.

Thus, the pressure of mixture of gases is

24.9 \times 10^{5} \text{Pa}

.

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