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  • Consider a sealed container with a volume of 0.04 . The container is filled with an ideal gas at a temperature of 300 K.

Consider a sealed container with a volume of 0.04 m^3. The container is filled with an ideal gas at a temperature of 300 K. The gas consists of 6.0 × 10^{23} gas molecules. Each molecule has a mass of 2.5 × 10^{-26} kg. Calculate the average force exerted by the gas molecules on the walls of the container, considering perfectly elastic collisions.

Option: 1

3.89 ×104 N
 


Option: 2

6.32 ×102 N


Option: 3

4.98 ×103 N
 


Option: 4

8.46 ×102 N


Answers (1)

best_answer

Given data:

                    Volume of the container (V ) = 0.04 m3
                                       Temperature (T) = 300 K
                        Number of molecules (N) = 6.0 × 1023
                     Mass of each molecule (m) = 2.5 × 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 300}{2.5 \times 10^{-26}}} \\ & \approx 436.77 \mathrm{~m} / \mathrm{s} \end{aligned}

Step 2: Calculate the average force exerted by the gas molecules on the walls of the container. The average force can be calculated using the formula:

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

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

\begin{aligned} F & =\frac{2 \times 6.0 \times 10^{23} \times 2.5 \times 10^{-26} \times(436.77)^2}{0.04} \\ & \approx 4.98 \times 10^3 \mathrm{~N} \end{aligned}

Therefore, the correct option is 1.

Posted by

Ritika Kankaria

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