An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm. And a temperature of `27^(@)C`. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm. And its temperature drops to `17^(@)C`. Estimate the mass of oxygen taken out of the cylinder. `(R = 8.1 J "mole"^(-1) K^(-1)`, molecular mass of `O_(2) = 32 u)`.

Explanation

Volume of oxygen, `V_(1)` = 30 litres `= 30 xx 10^(–3) m^3`
Gauge pressure, `P_(1) = 15 atm = 15 × 1.013 × 10^5 `Pa
Temperature, `T_(1) = 27^(@)C = 300 K`
Universal gas constant, `R = 8.314 J "mole"^(–1)" " K^(–1)`
Let the initial number of moles of oxygen gas in the cylinder be `n_(1)`.
The gas equation is given as:
` P_1V_1 = n_1RT_1
`therefore n_1=(P_1V_1)/(RT_1)`
`=(15.195xx10^5xx30xx10^(-3))/((8.314)xx300) `= 18.276
But, `n_1=(m_1)/(M)` Where,
`m_1 `= Initial mass of oxygen
M = Molecular mass of oxygen = 32 g
` therefore m_1 = n_1M = 18.276 xx 32 = 584.84 g`
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
Volume,` V_2 = 30 "litres" = 30 xx 10^(–3) m^3 `
Gauge pressure, `P_2 `= 11 atm = 11 x 1.013 × 105 Pa
Temperature, `T_2` = 17°C = 290 K
Let `n_2` be the number of moles of oxygen left in the cylinder.
The gas equation is given as:
`P_2V_2 = n_2RT_2 ` `therefore n_2=(P_2V_2)/(RT_2)`
`=(11.143xx10^5xx30xx10^(-3))/(8.314xx290)`
= 13.86
But,
Where,
`m_2` is the mass of oxygen remaining in the cylinder
`therefore m_2 = n_2 M = 13.86 xx 32 = 453.1 `g
The mass of oxygen taken out of the cylinder is given by the relation:
Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder
= `m_1 – m_2`
= 584.84 g – 453.1 g
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.