Q.An oxygen cylinder of volume $30$ litre has an initial gauge pressure of $15\ \text{atm}$ and a temperature of $27\ ^\circ\text{C}$. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11\ \text{atm}$ and its temperature drops to $17\ ^\circ\text{C}$. Estimate the mass of oxygen taken out of the cylinder ($R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$, molecular mass of $\text{O}_2 = 32\ \text{u}$).

Concept: Ideal Gas Law — the mass of gas is proportional to $PV/T$, so the mass removed is the difference between the initial and final masses.

Step 1: Convert to absolute pressure and temperature.

Gauge pressure = absolute pressure − 1 atm.

Initial: $P_1 = 15 + 1 = 16\ \text{atm}$, $T_1 = 27 + 273 = 300\ \text{K}$.

Final: $P_2 = 11 + 1 = 12\ \text{atm}$, $T_2 = 17 + 273 = 290\ \text{K}$.

Volume $V = 30\ \text{L} = 30 \times 10^{-3}\ \text{m}^3$.

Step 2: Find initial and final moles.

$n_1 = \frac{P_1 V}{R T_1}$, $n_2 = \frac{P_2 V}{R T_2}$.

Moles removed: $\Delta n = n_1 - n_2 = \frac{V}{R} \left( \frac{P_1}{T_1} - \frac{P_2}{T_2} \right)$.

Step 3: Compute numerically.

$\frac{P_1}{T_1} = \frac{16}{300} = 0.05333\ \text{atm/K}$,

$\frac{P_2}{T_2} = \frac{12}{290} = 0.04138\ \text{atm/K}$,

difference $= 0.01195\ \text{atm/K}$. …

The key to this problem is understanding that gauge pressure is not absolute pressure. A gauge reads zero when the cylinder is open to the atmosphere, so the actual pressure inside is gauge pressure plus atmospheric pressure (1 atm). The cylinder’s volume is fixed, so the number of moles changes with pressure and temperature according to the ideal gas law.

We also need to be careful with units: volume is in litres, pressure in atm, but $R$ is given in SI units (J/mol·K). We’ll convert everything to SI (m³, Pa, K) to keep the calculation clean.

1. Convert all quantities to SI units

   Volume: $V = 30\ \text{L} = 30 \times 10^{-3}\ \text{m}^3 = 0.030\ \text{m}^3$

   Temperatures:

   $T_1 = 27^\circ\text{C} = 27 + 273 = 300\ \text{K}$

   $T_2 = 17^\circ\text{C} = 17 + 273 = 290\ \text{K}$

   Pressures: Gauge pressure is relative to atmosphere. Absolute pressure = gauge + 1 atm.

   $1\ \text{atm} = 1.013 \times 10^5\ \text{Pa}$

   Initial absolute pressure: $P_1 = (15 + 1) \times 1.013 \times 10^5 = 16 \times 1.013 \times 10^5\ \text{Pa}$

   Final absolute pressure: $P_2 = (11 + 1) \times 1.013 \times 10^5 = 12 \times 1.013 \times 10^5\ \text{Pa}$

   Watch out

   A common mistake is to use gauge pressure directly in the ideal gas law. The gas law requires absolute pressure — always add 1 atm to the gauge reading.

2. Apply the ideal gas law to find initial and final moles

   The ideal gas law: $PV = nRT$

   For the initial state:

Let’s compute step by step:

$P_1 V = 16 \times 1.013 \times 10^5 \times 0.030 = 16 \times 1.013 \times 3000$ (since $10^5 \times 0.030 = 3000$)

$= 16 \times 3039 = 48,624\ \text{Pa·m}^3$ (which is Joules)

Denominator: $R T_1 = 8.31 \times 300 = 2493\ \text{J/mol}$

So $n_1 = \frac{48,624}{2493} \approx 19.51\ \text{mol}$

For the final state: …

A more direct route skips finding $n_1$ and $n_2$ separately: since $m=\frac{PVM}{RT}$ for an ideal gas, plug the initial and final $(P,T)$ pairs straight into this single formula to get $m_1$ and $m_2$, then subtract: $\Delta m=\frac{VM}{R}\left(\frac{P_1}{T_1}-\frac{P_2}{T_2}\right)$ — one substitution instead of a four-step mole-then-mass chain. It's also worth flagging as a standing check: gauge pressure m …