Problem 109 - A Complete Resource Book in Chemistry for JEE Main

Question

An L.P.G. cylinder contains \(15 \mathrm{~kg}\) of butane gas at \(27^{\circ} \mathrm{C}\) and 10 atmospheric pressure. It was leaking and its pressure fell down to 8 atmospheric pressure after one day. The gas leaked in 5 days is (a) \(10 \mathrm{~kg}\) (b) \(3 \mathrm{~kg}\) (c) \(15 \mathrm{~kg}\) (d) \(12 \mathrm{~kg}\)

Step-by-Step Solution

Answer

The gas leaked is 3 kg.

Step 1: Analyze Initial Conditions

The initial conditions of the LPG cylinder are given as \(15 \text{ kg}\) of butane gas at \(27^{\circ} \text{C}\) and a pressure of \(10 \text{ atm}\). These are the key parameters to consider for gauging changes in the gas amount due to leakage.

Step 2: Apply Ideal Gas Law Conditions

Use the Ideal Gas Law, \(PV = nRT\), to relate conditions before and after leakage. Initially, the amount (in moles) is \(n_1\) and the LPG cylinder pressure \(P_1 = 10 \text{ atm}\). After leakage, the pressure \(P_2 = 8 \text{ atm}\). The temperature and volume remain constant.

Step 3: Calculate Initial and Final Gas Amounts

Since \(T\) and \(V\) are constant, the ratio \(\frac{P_1}{n_1} = \frac{P_2}{n_2}\). Find \(n_1\) from initial gas mass: \(n_1 = \frac{15 \times 1000}{M} = \frac{15000}{58} \approx 258.6\) moles (where \(M = 58\) g/mol is the molar mass of butane). Calculate \(n_2: n_2 = \frac{8}{10} \times 258.6 \approx 206.9\) moles.

Step 4: Calculate Moles and Mass Leaked

The moles leaked is \(n_1 - n_2 = 258.6 - 206.9 = 51.7\) moles. Convert to mass leaked: \(51.7 \times 58 = 2998.6 \approx 3\) kg.

Key Concepts

LPG Cylinder

Liquefied Petroleum Gas (LPG) cylinders are common household utilities, providing fuel for cooking and heating. An LPG cylinder typically contains gases like propane or butane stored under pressure. The cylinder is constructed to handle high pressure to keep the gas in its liquid form. This storage method is not just energy-efficient but also ensures a compact way to transport and use fuel.
In our example, we are dealing with butane gas under specific pressure and temperature conditions. The study of this setup helps understand how gas behaves when external variables like pressure change, which in turn affects gas amount due to leaks.

Butane Gas

Butane is a type of alkane with the chemical formula C₄H₁₀, commonly found in LPG cylinders. It is colorless, easily liquefied gas used for fuel and industrial processes. One of butane's essential properties is its molar mass of 58 g/mol, which is crucial for calculations involving gas laws.
The behavior of butane under pressure is often analyzed with the Ideal Gas Law, which helps in understanding how temperature and pressure influence the volume and mass of butane in practical scenarios. This is exactly what we observed in the problem, where we need to account for changes in butane due to leakage.

Pressure Changes

Pressure is a key factor in determining the state and behavior of gases within a closed system, such as an LPG cylinder. Initially, the pressure inside the cylinder is 10 atmospheric pressures, indicating a tightly packed gas. If there's a leak, pressure drops, indicating a loss of gas. This change is pivotal in calculating how much gas remains or is lost.
The Ideal Gas Law \[PV = nRT\] becomes useful as it connects various parameters like pressure \(P\), volume \(V\), and number of moles \(n\). When pressure decreases, it reflects in a decrease in the number of moles, assuming temperature and volume remain constant. This relationship helps determine the quantity of gas leaked in the described scenario.

Molar Mass

Molar mass is the mass of one mole of a substance and is key in transitioning between the mass of a gas and its moles during calculations. In the context of butane, with a molar mass of 58 g/mol, this property helps convert between the physical weight of the gas and the amount in moles.
For example, when you have 15 kg of butane in the cylinder, to find out its moles, you divide the mass by the molar mass: \[ n_1 = \frac{15 \times 1000}{58} = 258.6 \text{ moles} \]
This calculation acts as a starting point to assess subsequent changes in the state, like leakage, in terms of mass rather than just volume or pressure.

Gas Leakage Calculations

Understanding gas leaks involves determining how much gas has escaped from the system. In our scenario, the pressure decrease from 10 to 8 atm allows us to calculate this change using the Ideal Gas Law.
Initially, you determine the moles of gas before and after the leak:\[ n_1 = 258.6 \text{ moles initially} \]
\[ n_2 = \frac{8}{10} \times 258.6 = 206.9 \text{ moles after leak} \]
The difference gives the moles of gas lost, converted to mass using the molar mass:\[ n_1 - n_2 = 51.7 \text{ moles} \]
\[ \text{Mass leaked} = 51.7 \times 58 \approx 3 \text{ kg} \]
Through these calculations, we find the result to be part of a critical process in ensuring the safe and efficient use of LPG systems.