If the temperature and pressure of a gas is doubled the mean free path of the gas molecules

Explanation

We have,
` dV/dt = r `
` rArr dV = rdt `
Let the pressure pumped out gas = dp `
` Volume of container = `V_0`
At a pump dV amount of gas has been
pumped out
` PdV = -(V_0) dP `
` rArr Prdt = - (V_0) dP`
` rArr dp/P = (- rdt / V_0) `
On integration , we get
` P = (e^ (-rt/V_0))`
` Half of the gas been pumped out , pressure
will be half
` i.e. 1 = (1/2 e ^ (-rt/ V_0))`
` rArr ` 1 n 2 = rt/ V_0 `
` rArr t = In 2 xx (V_0/r )`