Oxygen is filled in a closed metal jar of volume `1.0xx10^(-3)m^(3)` at a pressure of `1.5xx10^(5)Pa.` and temperature 400K.The jar has a small leak in it. The atmospheric pressure is `1.0xx10^(5)`Pa and the atmospheric temperature is 300K. Find the mass of the gas that leaks out by time the pressure and the temperature inside the jar equlise with the surrounding.

Explanation

Given
` (V_1)=(1 xx (10^-3)(m^3))`
` P_1= (1.5 xx (10^5)Pa)`
` T_1= 400K`
We know
` (P_1)(V_1)= (n_1)(RT_1)`
` rArr n_1 = ((P_1)(V_1)/RT_1)`
` =(1.5 xx (10^5)xx 1 xx (10^-3)/(8.3 xx 400))`
` rArr (m_1) = (1.5 xx M/8.3 xx 4) = (1.5 xx 32/ 4 xx 8.3)`
` = 1.457 ~~1.446`
` Again, (P_2)= 1 xx (10^5)Pa,`
` V_2 = 1 xx (10^-3)(m^3)`
` T_2 = 300K`
We know,
` (P_2)(V_2)= (n_2)RT_2`
` rArr (n_2)= ((P_2)(V_2)/(RT_2)) = (10^5 xx (10^-3)/ (8.3 xx 300))`
` = (1/3 xx 8.3)= 0.040`
` rArr m_2= 0.04 xx 32 = 1.285`
` Delta m = (m_1-(m_2))=1.446 -1.285`
` = 0.1608 g ~~ 0.16g.`