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Consider a sample of gas in a fixed volume container . From the arguments of Kinetic Theory of Gases on quadrupling the temperature what will be the effect on pressure .

Basically I thought that the answer should be that the pressure gets doubled . But the answer is that it gets quadrupled . Can anyone explain how ??

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  • $\begingroup$ Why do you think it should result in a two fold increase in pressure? $\endgroup$ – Nanoputian Oct 10 '15 at 5:23
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Consider the following diagram of a particle. Now under the assumptions from kinetic theory, the average force on the walls of the container is given by: $$F_{average} = \frac{mNv^2_x}{L}$$ Now assuming that the speed of the particle is random in all directions we get: $$v^2 = v^2_x + v^2_y + v^2_z = 3v^2_x$$ Now pressure is defined as the average force acting on a surface area of $\mathrm{1m^2}$. Therefore the pressure is: $$p = \frac{F_{avg}}{A} = \frac{mNv^2}{3LA} = \frac{mNv^2}{3V} = \frac{2N}{3V}\frac{1}{2}mv^2$$ So we can see that the pressure on the wall is directly proportional to $1/2mv^2$. This relates to the total kinetic energy of the particle. Temperature is defined as the average kinetic energy of a particle, hence $1/2mv^2$ is directly proportional to temperature. Therefore a two fold increase in temperature will result in a two fold increase in kinetic energy which will result in a four two increase in pressure.

Hence $p \propto T$. Therefore a four fold increase in temperature will result in a four fold increase in pressure.

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  • $\begingroup$ Prove it with Kinetic theory of gases $\endgroup$ – Tejus Oct 10 '15 at 5:58
  • $\begingroup$ Sorry I misunderstood your question. I will edit my answer so that it uses Kinetic theory of gases. $\endgroup$ – Nanoputian Oct 10 '15 at 6:08

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