A vessel of volume `V = 30 1` contains ideal gas at the temperature `0 ^@ C`. After a portion of the gas has been let out, the pressure in the vessel decreased by `Delta p = 0.78 atm` (the temperature remaining constant). Find the mass of the released gas. The gas density under the normal conditions `rho = 1.3 g//1`.

Explanation

Let `m_1` and `m_2` be the masses of the gas in the vessel before and after the gas is released.
Hence mass of the gas released,
`Delta m = m_1 - m_2`
Now from ideal gas equation
`p_1 V = m_1 ( R)/(M) T_0` and `p_2 V = m_2 ( R)/(M) T_0`
as `V` as `T` are same before and after the release of the gas.
so, `(p_1 - p_2) V = (m_1 - m_2) ( R)/(M)T_0 = Delta m ( R)/(M) T_0`
or, `Delta m=((p_1 - p_2)VM)/(R T_0) = (Delta p VM)/(R T_0)`....(1)
We also know `p = rho (R)/(M) T so, (M)/(R T_0) = (rho)/(p_0)`...(2)
(where `p_0` = standard atmospheric pressure and `T_0 = 273 K`)
From Eqs. (1) and (2) we get
`Delta m = rho V (Delta p)/(p_0) = 1.3 xx 30 xx (0.78)/(1) = 30 g`.