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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 $6.0 \times 10^{23}$ gas molecules. Each molecule has a mass of $2.5 \times 10^{-26}$ kg. Calculate the average force exerted by the gas molecules on the walls of the container, considering perfectly elastic collisions.
Option: 1
$3.89\times 10^4N$

Option: 2
$6.32\times 10^2N$

Option: 3
$4.98\times 10^3N$

Option: 4
$846\times 10^2N$

Answers (1)

best_answer

Given data:

Volume of the container (V ) = 0.04 m3
Temperature (T) = 300 K
Number of molecules (N) $=6.0\times10^{23}$

Mass of each molecule (m) $=2.5\times10^{-26}kg$

Boltzmann constant (k) $=1.38\times10^{-23}J/K$

Step 1: Calculate the root mean square (rms) speed of the gas molecules. The rms speed is given by the formula:

$v_{rms}=\sqrt{\frac{3kT}{m}}$

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

$v_{rms}= \sqrt{\frac{3\times1.38\times10^{-23}\times 300}{2.5\times10^{-26}}}$

$\approx 436.77 m/s$

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:

$F= \frac{2mv^2_{rms}}{V}$

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

$F=\frac{2\times6.0\times10^{23}\times2.5\times10^{-26}\times (436.77)^2}{0.04}$

$\approx 4.98\times10^3N$

Therefore, the correct option is 3.

Posted by

Riya

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