Question

The pressure exerted by $6.0 \, g$ of methane gas in a $0.03\, m^3$ vessel at $129^{\circ}C$ is (Atomic masses : $C = 12.01, H = 1.01 $ and $R = 8.314 \, JK^{-1} mol^{-1}$)

• A. 13409 Pa
• B. 41648 Pa
• C. 31684 Pa
• D. 215216 Pa

Answer 1

The Correct Option is B

 To find the pressure exerted by 6.0 g of methane gas in a 0.03 m³ vessel at 129°C, we can use the ideal gas equation:

\(PV = nRT\)

where:

• \(P\) is the pressure in Pascals (Pa)
• \(V = 0.03 \, m^3\) is the volume of the container
• \(n\) is the number of moles of gas
• \(R = 8.314 \, JK^{-1}mol^{-1}\) is the ideal gas constant
• \(T\) is the temperature in Kelvin (K)

First, we need to convert the temperature from Celsius to Kelvin:

\(T = 129^{\circ}C + 273.15 = 402.15 \, K\)

Next, calculate the number of moles \((n)\) of methane gas. The molar mass of methane \((CH_4)\) is:

\(M = 12.01 + 4 \times 1.01 = 16.05 \, g/mol\)

Therefore, the number of moles is:

\(n = \frac{6.0 \, g}{16.05 \, g/mol} \approx 0.3736 \, mol\)

Now, we can substitute these values into the ideal gas equation to find the pressure \((P)\):

\(P \cdot 0.03 = 0.3736 \times 8.314 \times 402.15\)

Solving for \(P\), we get:

\(P = \frac{0.3736 \times 8.314 \times 402.15}{0.03}\)

Calculate the right-hand side:

\(P \approx 41648 \, Pa\)

Thus, the pressure exerted by the methane gas is 41648 Pa.

This matches the correct option:

41648 Pa

.