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A gas is confined to a cubic container with sides of length 0.05 m. The temperature of the gas is 500 K, and each gas molecule has a mass of 2.8 × 10−26 kg. Calculate the average force exerted by the gas molecules on the walls of the container and determine the pressure inside the container.

Option: 1

1.458 ×10−10 Pa


Option: 2

1.785 ×10−15 Pa


Option: 3

2.158 ×10−13 Pa


Option: 4

2.562 ×10−18 Pa


Answers (1)

best_answer

Given data: Side length of the container (L) = 0.05 m

Temperature (T) = 500 K

Mass of each gas molecule (m) = 2.8 × 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 500}{\pi \times 2.8 \times 10^{-26}}} \\ \langle v\rangle \approx 603.62 \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 2.8 \times 10^{-26} \times N \times(603.62)^2 \times \frac{1}{0.05}F_{\text {avg }} \approx 6.472 \times 10^{-15} \mathrm{~N}

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

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

where A is the area of one of the container’s walls. Since the container is cubic, A = 6L2 (total surface area). Substitute the values of Favg and L and solve for P:

\begin{gathered} P=\frac{6.472 \times 10^{-15}}{6 \times(0.05)^2} \\ P \approx 2.158 \times 10^{-13} \mathrm{~Pa} \end{gathered}

Therefore, the option correct is (3). article amsmath

 

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rishi.raj

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