Force acting on the area of the cylinder due to gas

pV=nRT

The Attempt at a Solution

I know the volume of the cylinder, which is Al. So I plugged this into the ideal gas law formula, and got answer B. However, the correct answer should be D. I see the Boltzmann's constant there in the equation, and I do know an equation which contains the constant, but I'm not sure how to incorporate it into the ideal gas law.. Thanks for your help in advance!

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PKM
Can you somehow relate ##P## with the kinetic energy of the gas molecules per unit volume ##E_{\text{unit vol.}}##?

dRic2
Gold Member
First, you need a Force, not a pressure. What you did, I think, was $$p = nRT/V = \frac {nRT} {A l}$$ but this is a pressure. You have to multiply by the area, so $$F = \frac {nRT} {l}$$.

Second: in the perfect gas law you have ##n## which is the number of moles, but the exercise gives you ##N## which is number of molecules! And Remember that ##k_B = \frac R N_a## where ##N_a## is the Avogadro constant

First, you need a Force, not a pressure. What you did, I think, was $$p = nRT/V = \frac {nRT} {A l}$$ but this is a pressure. You have to multiply by the area, so $$F = \frac {nRT} {l}$$.

Second: in the perfect gas law you have ##n## which is the number of moles, but the exercise gives you ##N## which is number of molecules! And Remember that ##k_B = \frac R N_a## where ##N_a## is the Avogadro constant
Thank you so much! I figured it out :)