Question

A cylinder contains hydrogen gas at pressure of 249 kPa and temperature $27^\circ\,C$. Its density is :$(R = 8.3\,J\,mol^{-1}K^{-1}$)

• A. $0.5 \,kg/m^3$
• B. $0.2 \,kg/m^3$
• C. $0.1 \,kg/m^3$
• D. $0.02 \,kg/m^3$

Answer 1

The Correct Option is B

To determine the density of hydrogen gas inside the cylinder, we can use the Ideal Gas Law equation:

\(PV = nRT\)

Where:

• \(P\) = Pressure of the gas in Pa (1 kPa = 1000 Pa)
• \(V\) = Volume of the gas in cubic meters
• \(n\) = Number of moles of the gas
• \(R\) = Universal gas constant = \(8.3 \,J \,mol^{-1}K^{-1}\)
• \(T\) = Temperature in Kelvin

First, let's convert the temperature from Celsius to Kelvin:

\(T = 27^\circ\,C + 273.15 = 300.15 \,K\)

Next, convert the pressure from kPa to Pa:

\(P = 249 \,kPa = 249,000 \,Pa\)

Rearrange the Ideal Gas Law to solve for the number of moles \(n\):

\(n = \frac{PV}{RT}\)

Using the relation \(n = \frac{mass}{molar \, mass}\), we can express density \(\rho\) as:

\(\rho = \frac{mass}{Volume} = \frac{n \times molar \, mass}{V} = \frac{P \times molar \, mass}{RT}\)

The molar mass of hydrogen (H₂) is approximately \(2 \,g/mol = 0.002 \,kg/mol\).

Substitute the values into the formula:

\(\rho = \frac{249,000 \times 0.002}{8.3 \times 300.15}\)

Calculate the density:

\(\rho = \frac{498}{2494.245} \approx 0.1997 \,kg/m^3 \approx 0.2 \,kg/m^3\)

Therefore, the correct answer is:

\(0.2 \,kg/m^3\)

This corresponds to the given correct answer option.