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Thermodynamics Part 2

Section: Physics

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Question : 10 of 100

Marks: +1, -0

The volume

V

of an enclosure contains a mixture of three gases,

16\,\text{g}

28\,\text{g}

of nitrogen and

44\,\text{g}

of carbon dioxide at absolute temperature

T

R

as universal gas constant. The pressure of the mixture of gases is

[16 Mar 2021 Shift 1]

$\;\frac{88 R T}{V}$
$\;\frac{3 R T}{V}$
$\;\frac{5}{2} \;\frac{R T}{V}$
$\;\frac{4 R T}{V}$

Solution:

From ideal gas equation,

p V=(n_{1}+n_{2}+n_{3}) R T

where,

n_{1}=

number of moles of oxygen

=\;\frac{16}{32}

n_{2}=

number of moles of nitrogen

=\;\frac{28}{28}

n_{3}=

number of moles of carbon dioxide

=\;\frac{44}{44}

\Rightarrow \;\; p V=\left[\;\frac{16}{32}+\;\frac{28}{28}+\;\frac{44}{44}\right] R T=\left[\;\frac{1}{2}+1+1\right]_{R T}

=\;\frac{5}{2} R T \Rightarrow p=\;\frac{5}{2} \;\frac{R T}{V}

This is the required pressure of the mixture of the gases.

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