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  • A closed container holds a certain quantity of an ideal gas. The container’s volume is 0.02 m3 and the temperature of the gas is 300 K. Each gas molecule has a mass of 3.2 ×

A closed container holds a certain quantity of an ideal gas. The container’s volume is 0.02 m3 and the temperature of the gas is 300 K. Each gas molecule has a mass of 3.2 × 10−26 kg. Calculate the average force exerted by the gas molecules on the walls of the container and also determine the pressure inside the container.

Option: 1

2.456 ×10−11 Pa


Option: 2

1.241 ×10−16 Pa


Option: 3

0.478 ×10−18 Pa


Option: 4

1.343 ×10−14 Pa


Answers (1)

best_answer

Given data: Volume of the container (V ) = 0.02 m3

Temperature (T) = 300 K

Mass of each gas molecule (m) = 3.2 × 10−26 kg

Boltzmann constant (k) = 1.38 × 10−23 J/K

Step 1: Calculate the average velocity of the gas molecules using the formula:

\langle v\rangle=\sqrt{\frac{8 k T}{\pi m}}

Substitute the given values and solve for ⟨v⟩:

\begin{gathered} \langle v\rangle=\sqrt{\frac{8 \times 1.38 \times 10^{-23} \times 300}{\pi \times 3.2 \times 10^{-26}}} \\ \langle v\rangle \approx 502.38 \mathrm{~m} / \mathrm{s} \end{gathered}

Step 2: Calculate the average force (Favg) exerted by the gas molecules on the walls of the container using the formula:

F_{\text {avg }}=2 m N\langle v\rangle^2 L^{-1}

Substitute the given values and solve for Favg:

F_{\text {avg }}=2 \times 3.2 \times 10^{-26} \times N \times(502.38)^2 \times \frac{1}{0.02}F_{\text {avg }} \approx 1.614 \times 10^{-16} \mathrm{~N}

Step 3: Calculate the pressure (P) inside the container using the formula:

P=\frac{F_{\mathrm{avg}}}{A}

where A is the area of one of the container’s walls. Since the container is closed and has six walls, A = 6V (total surface area). Substitute the values of Favg and V and solve for P:

P=\frac{1.614 \times 10^{-16}}{6 \times 0.02}

P \approx 1.343 \times 10^{-14} \mathrm{~Pa}

Therefore, the option correct is (4). article amsmath

 

Posted by

Devendra Khairwa

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