Question

An LPG (Liquified Petroleum Gas) cylinder weighs 15.0 kg when empty. When full, it weighs 30.0 kg and shows a pressure of 3.0 atm. In the course of usage at 27°C, the mass of the full cylinder is reduced to 24.2 kg. The volume of the used gas in cubic metre at the normal usage condition (1 atm and 27°C) is (assume LPG to be normal butane and it behaves ideally):

• A. 24.6 m\textsuperscript{3
• B. 246 m\textsuperscript{3
• C. 0.246 m\textsuperscript{3
• D. 2.46 m\textsuperscript{3

Hint:

Tip on gas law problems In gas law problems, remember to always convert mass to moles, use the ideal gas law for calculating volume, and make sure all units are consistent. For gases, 1 mole of an ideal gas occupies 22.4 L at STP, but adjustments need to be made for non-standard conditions.

Answer 1

The Correct Option is D

Given:
- Mass of gas in cylinder when full: \(30.0 \, \text{kg}\)
- Mass of gas remaining in the cylinder: \(24.2 \, \text{kg}\)
- Pressure of gas when full: \(3.0 \, \text{atm}\)
- Temperature during usage: \(27^\circ \text{C}\)
To find the volume of the used gas at normal usage conditions, we use the ideal gas law: \[ PV = nRT \] Steps: First, calculate the moles of butane gas used: \[ \text{Mass of gas used} = 30.0 \, \text{kg} - 24.2 \, \text{kg} = 5.8 \, \text{kg} \] Molar mass of butane (C\textsubscript{4}H\textsubscript{10}) = 58 g/mol = 0.058 kg/mol.
Therefore, the number of moles of butane used is: \[ \nn = \frac{\text{Mass of gas used}}{\text{Molar mass}} = \frac{5.8}{0.058} = 100 \, \text{moles} \] Now, use the ideal gas law to calculate the volume at standard conditions (1 atm, 27°C = 300 K): \[ \nV = \frac{nRT}{P} \] Where: - \(R = 0.0821 \, \text{L·atm/mol·K}\) (the gas constant) - \(T = 300 \, \text{K}\) (temperature in Kelvin) - \(P = 1 \, \text{atm}\) (pressure at normal usage) Substitute the values: \[ \nV = \frac{(100 \, \text{mol})(0.0821 \, \text{L·atm/mol·K})(300 \, \text{K})}{1 \, \text{atm}} = 2460 \, \text{L} = 2.46 \, \text{m}^3 \] Hence, the volume of the used gas is \(2.46 \, \text{m}^3\).