Question

A cylinder contains 20n kg of ${\mathrm{N}}_2$ gas $(\mathrm{M}=28\mathrm{kg}{\mathrm{k}}^{-1}{\mathrm{mol}}^{-1})$ at a pressure of 5 atm. The mass of hydrogen $(\mathrm{M}=2\mathrm{kg}{\mathrm{k}}^{-1}{\mathrm{mol}}^{-1})$ at a pressure of 3 atm contained in the same cylinder at the same temperature is:

• A. 1.08 kg
• B. 0.86 kg
• C. 0.68 kg
• D. 1.68 kg

Answer 1

The correct option is B

0.86 kg

The explanation for the correct answer:-

Option (B) 0.86 kg

Step 1: Given data

Mass of nitrogen given$\left({\mathrm{m}}_1\right)$ = $20\mathrm{kg}$

The molecular mass of nitrogen$\left({\mathrm{M}}_1\right)=28\mathrm{kg}{\mathrm{k}}^{-1}{\mathrm{mol}}^{-1}$

Pressure in case of Nitrogen $\left({\mathrm{P}}_1\right)=5\mathrm{atm}$

The molecular mass of hydrogen$\left({\mathrm{M}}_2\right)=2\mathrm{kg}{\mathrm{k}}^{-1}{\mathrm{mol}}^{-1}$

Pressure in case of Nitrogen $\left({\mathrm{P}}_2\right)=3\mathrm{atm}$

Step 2: Finding the relation between the two gases

We know that, for an Ideal gas,

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \Rightarrow \frac{\mathrm{V}}{\mathrm{T}}=\frac{\mathrm{nR}}{\mathrm{P}} \end{gathered}$$

Here V= Volume of the gas, T= Temperature of the gas, R= Universal gas constant, P= Pressure of the gas, n= number of moles of the gas

Let the mass of hydrogen be $m_2$

So for Nitrogen, $$\begin{gathered} \frac{\mathrm{V}}{\mathrm{T}}=\frac{{\mathrm{n}}_1\mathrm{R}}{{\mathrm{P}}_1} \\ \Rightarrow \frac{\mathrm{V}}{\mathrm{T}}=\frac{{\mathrm{m}}_1\mathrm{R}}{{\mathrm{P}}_1{\mathrm{M}}_1} \end{gathered}$$and for Hydrogen $$\begin{gathered} \frac{\mathrm{V}}{\mathrm{T}}=\frac{{\mathrm{n}}_2\mathrm{R}}{{\mathrm{P}}_2} \\ \Rightarrow \frac{\mathrm{V}}{\mathrm{T}}=\frac{{\mathrm{m}}_2\mathrm{R}}{{\mathrm{P}}_2{\mathrm{M}}_2} \end{gathered}$$ $\left[\because \mathrm{Number}\mathrm{of}\mathrm{moles}\left(\mathrm{n}\right)=\frac{\mathrm{Given}\mathrm{mass}\left(\mathrm{m}\right)}{\mathrm{Molecular}\mathrm{mass}\left(\mathrm{M}\right)}\right]$

So from the above equations, we can write as:

$\frac{{\mathrm{m}}_1\mathrm{R}}{{\mathrm{P}}_1{\mathrm{M}}_1}=\frac{{\mathrm{m}}_2\mathrm{R}}{{\mathrm{P}}_2{\mathrm{M}}_2}$

Since there are two different gases in the same container, the Volume is the same for both gases, also given that the temperature is the same for both as well.

Step 3: Finding the mass of hydrogen

We have,

$$\begin{gathered} \frac{{\mathrm{m}}_1\mathrm{R}}{{\mathrm{P}}_1{\mathrm{M}}_1}=\frac{{\mathrm{m}}_2\mathrm{R}}{{\mathrm{P}}_2{\mathrm{M}}_2} \\ \Rightarrow {\mathrm{m}}_2=\frac{{\mathrm{m}}_1{\mathrm{P}}_2{\mathrm{M}}_2}{{\mathrm{P}}_1{\mathrm{M}}_1} \end{gathered}$$

Putting all the given value we get,

$$\begin{gathered} {\mathrm{m}}_2=\frac{{\mathrm{m}}_1{\mathrm{P}}_2{\mathrm{M}}_2}{{\mathrm{P}}_1{\mathrm{M}}_1} \\ \Rightarrow {\mathrm{m}}_2=\frac{20\times 3\times 2}{5\times 28} \\ \Rightarrow {\mathrm{m}}_2=\frac{20\times 3\times 2}{5\times 28} \\ \Rightarrow {\mathrm{m}}_2=0.8571 \\ \Rightarrow {\mathrm{m}}_2\approx 0.86\mathrm{kg} \end{gathered}$$

Therefore, the mass of hydrogen is $0.86\mathrm{kg}$, option (B) is the correct answer.