Question

A gas in a cylinder is at pressure P. If the masses of all the molecules are made one third of their original value and their speeds are doubled, then find the resultant pressure.

Answer 1

Given: m₂ = m₁/3 v₂ = 2v₁

Pressure P= \(\frac13 \frac{mN}{v}v^2\)

∴P₁= \(\frac13 \frac{m_1N}{v}v_1^2\)

∴P₂= \(\frac13 \frac{m_2N}{v}v_2^2\)

∴ \(\frac{P_1}{P_2} =(\frac{m_2}{m_1})(\frac{v_2^2}{v_1^2})\)

= \(\frac{\frac{m_1}{3}}{m_1}(2^2)\) = \(\frac43\)

P₂ = \(\frac43\) P₁