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Consider a closed container with a volume of 0.010 \mathrm{~m}^{3}, filled with helium gas (He) at a temperature of 400 \mathrm{~K}. The gas molecules in the container are assumed to follow the ideal gas behavior. Calculate the average force exerted by 2 mole helium molecules on the container walls during a collision.

Option: 1

300 \mathrm{~N}


Option: 2

321.63 \mathrm{~N}


Option: 3

400 \mathrm{~N}


Option: 4

440 \mathrm{~N}


Answers (1)

Step 1: Calculate the root mean square (RMS) velocity \left(v_{r m s}\right) using the formula:v_{r m s}=\sqrt{\frac{3 k T}{m}}where: k = Boltzmann constant \left(1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right) T= temperature in Kelvin $-m=$ mass of one molecule in \mathrm{kg}

Solution:

\begin{gathered} v_{r m s}=\sqrt{\frac{3 \cdot\left(1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right) \cdot 400 \mathrm{~K}}{6.646 \times 10^{-27} \mathrm{~kg}}} \\ v_{r m s} \approx 1796.6 \mathrm{~m} / \mathrm{s} \end{gathered}

Step 2: Calculate the average force F exerted by a single helium molecule on the container walls during a collision.

F=\frac{2 m v_{x}}{t}

where: m= mass of the molecule v_{x}=\mathrm{x} component of the velocity t= time for a 'round trip', which is t=\frac{2 L}{v_{x}}, where $L$ is the dimension of the container

Solution:

Assuming the container has a dimension L in the x-direction, let's assume L=0.01 \mathrm{~m}.

t=\frac{2 L}{v_{x}}=\frac{2 \times 0.01 \mathrm{~m}}{1796.6 \mathrm{~m} / \mathrm{s}} \approx 1.11 \times 10^{-5} \mathrm{~s}

Now, we can calculate the average force using the formula:

F=\frac{2 m \cdot v_{x}}{t}=\frac{2 \times 6.646 \times 10^{-27} \mathrm{~kg} \cdot 1796.6 \mathrm{~m} / \mathrm{s}}{1.11 \times 10^{-5} \mathrm{~s}} \approx 2.67 \times 10^{-22} \mathrm{~N}
 

Step 3: Calculate the total force exerted by all helium molecules on the container walls. 

\begin{gathered} F=2.67 \times 10^{-22} \times 6.023 \times 10^{23} \\ F=321.63 N \end{gathered}

Posted by

Ramraj Saini

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