A cylindrical container of volume

Question:

A cylindrical container of volume $4.0 \times 10^{-3} \mathrm{~m}^{3}$ contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is $400 \mathrm{~K}$. The pressure of the mixture of gases is :

[Take gas constant as $8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ]

 

  1. $249 \times 10^{1} \mathrm{~Pa}$

  2. $24.9 \times 10^{3} \mathrm{~Pa}$

  3. $24.9 \times 10^{5} \mathrm{~Pa}$

  4. $24.9 \mathrm{~Pa}$


Correct Option: , 3

Solution:

$\mathrm{V}=4 \times 10^{-3} \mathrm{~m}^{3}$

$\mathrm{n}=3$ moles

$\mathrm{T}=400 \mathrm{~K}$

$\mathrm{PV}=\mathrm{nRT} \Rightarrow \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}$

$P=\frac{3 \times 8.3 \times 400}{4 \times 10^{-3}}$

$=24.9 \times 10^{5} \mathrm{~Pa}$

Ans 3

 

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