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  • A cubic container with sides of length 0.1 m is filled with an ideal gas at a temperature of 600 K. The gas contains 5.0×<img alt="10^{24}" src="https://entrancecorner.oncodecogs.com/gif.late

A cubic container with sides of length 0.1 m is filled with an ideal gas at a temperature of 600 K. The gas contains 5.0×10^{24} gas molecules. Each molecule has a mass of 2.0 × 10^{-26} kg. Determine the root mean square speed of the gas molecules, the total kinetic energy of the gas, and the pressure exerted by the gas on the walls of the container. All the given assumptions apply.

Option: 1

2.30 \times 10^9 \mathrm{~N} / \mathrm{m}^2


Option: 2

1.39 \times 10^4 \mathrm{~N} / \mathrm{m}^2


Option: 3

2.53 \times 10^3 \mathrm{~N} / \mathrm{m}^2


Option: 4

2.38 \times 10^5 \mathrm{~N} / \mathrm{m}^2


Answers (1)

Given data:
Length of each side of the cubic container (L) = 0.1 m
                                               Temperature (T) = 600 K
                          Number of molecules (N) = 5.0 × 10^{24}
                   Mass of each molecule (m) = 2.0 × 10^{-26} kg
                   Boltzmann constant (k) = 1.38 × 10^{-23} J/K
Step 1: Calculate the root mean square (rms) speed of the gas molecules. The rms speed is given by the formula:

v_{\mathrm{rms}}=\sqrt{\frac{3 k T}{m}}

Substitute the given values and solve for v_{rms}:

\begin{aligned} v_{\mathrm{rms}} & =\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 600}{2.0 \times 10^{-26}}} \\ & \approx 1592.69 \mathrm{~m} / \mathrm{s} \end{aligned}

Step 2: Calculate the total kinetic energy (KE) of the gas. The total
kinetic energy is given by:

K E=\frac{3}{2} N k T
Substitute the given values and solve for KE:

\begin{aligned} K E & =\frac{3}{2} \times 5.0 \times 10^{24} \times 1.38 \times 10^{-23} \times 600 \\ & \approx 2.07 \times 10^{-15} \mathrm{~J} \end{aligned}

Step 3: Calculate the pressure (P) exerted by the gas on the walls of the
container. The pressure can be calculated using the formula:

P=\frac{N m v_{\mathrm{rms}}^2}{3 V}

Substitute the given values and the calculated vrms, and solve for P:

\begin{aligned} P & =\frac{5.0 \times 10^{24} \times 2.0 \times 10^{-26} \times(1592.69)^2}{3 \times(0.1)^3} \\ & \approx 2.38 \times 10^5 \mathrm{~N} / \mathrm{m}^2 \end{aligned}

Therefore, the correct option is 4.

Posted by

Ramraj Saini

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