JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 18)
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 :
$${{3RT} \over V}$$
$${{4RT} \over V}$$
$${{88RT} \over V}$$
$${5 \over 2}{{RT} \over V}$$
Explanation
No. of moles of O2 :
n1 = $${{16} \over {32}}$$ = 0.5 mole
No. of moles of N2 :
n2 = $${{28} \over {28}}$$ = 1 mole
No. of moles of CO2 :
n3 = $${{44} \over {44}}$$ = 1 mole
Total no. of moles in container : n = n1 + n2 + n3
$$ \therefore $$ n = 0.5 + 1 + 1 = $${5 \over 2}$$ moles
Now; PV = nRT
$$ \Rightarrow $$ P = $${{nRT} \over V}$$
$$ \Rightarrow $$ P = $${5 \over 2}{{RT} \over V}$$
n1 = $${{16} \over {32}}$$ = 0.5 mole
No. of moles of N2 :
n2 = $${{28} \over {28}}$$ = 1 mole
No. of moles of CO2 :
n3 = $${{44} \over {44}}$$ = 1 mole
Total no. of moles in container : n = n1 + n2 + n3
$$ \therefore $$ n = 0.5 + 1 + 1 = $${5 \over 2}$$ moles
Now; PV = nRT
$$ \Rightarrow $$ P = $${{nRT} \over V}$$
$$ \Rightarrow $$ P = $${5 \over 2}{{RT} \over V}$$
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