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 $8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ]
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
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.