The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :
Solution
No. of moles of O<sub>2</sub> :
<br>n<sub>1</sub> = ${{16} \over {32}}$ = 0.5 mole<br><br>No. of moles of N<sub>2</sub> :
<br>n<sub>2</sub> = ${{28} \over {28}}$ = 1 mole<br><br>No. of moles of CO<sub>2</sub> :
<br>n<sub>3</sub> = ${{44} \over {44}}$ = 1 mole<br><br>Total no. of moles in container : n = n<sub>1</sub> + n<sub>2</sub> + n<sub>3</sub><br><br>$\therefore$ n = 0.5 + 1 + 1 = ${5 \over 2}$ moles<br><br>Now; PV = nRT<br><br>$\Rightarrow$ P = ${{nRT} \over V}$<br><br>$\Rightarrow$ P = ${5 \over 2}{{RT} \over V}$
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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