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Chapter 3: Problem 109

An L.PG. 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) \(12 \mathrm{~kg}\)

Short Answer

Expert verified
The gas leaked in 5 days is 3 kg, so the answer is (b).

Step by step solution

01

Understand the Initial Situation

We start with a cylinder containing 15 kg of butane at a temperature of \(27^{\circ} \mathrm{C}\) (which is \(300.15\, K\) when converted to Kelvin) and a pressure of \(10 \text{ atm}\).
02

Understand the Situation After Leak

After one day, the pressure inside the cylinder decreases to \(8 \text{ atm}\) while the temperature remains constant.
03

Apply the Ideal Gas Law for Initial Condition

For the initial condition: \[ PV = nRT \] where \( P = 10 \, \text{atm} \), \( V\) is volume, \( n_1 \) is moles of gas, \( R\) is the ideal gas constant and \( T = 300.15 \, K \). Since butane has a molar mass of \(58.12 \, g/mol\), the initial moles \( n_1 \) can be calculated using \( 15 \, kg \) converted to grams \( 15000 \, g\): \[ n_1 = \frac{15000}{58.12} \].
04

Calculate the Moles After Leak

After the leak, the pressure is \(8 \, \text{atm}\). We use the same volume and temperature, hence: \[ PV = n_2RT \] Substitute \( P = 8 \, \text{atm} \) and solve for final moles \( n_2 \): \[ n_2V = \frac{8}{10}n_1V \] Thus, \( n_2 = 0.8n_1 \).
05

Calculate the Mass of Gas Leaked

Using \( n = \frac{m}{M} \) (where \( m \) is the mass and \( M \) is the molar mass), calculate the mass corresponding to \( n_2 \) and \( n_1 \):\[ m_2 = n_2 \times M = 0.8 \times n_1 \times M = 0.8 \times 15000 \, g = 12000 \, g = 12 \, kg \]Therefore, the mass of gas leaked is \( m_1 - m_2 = 15 \, kg - 12 \, kg = 3 \, kg \).
06

Conclude

After calculating, we find the mass of the butane gas leaked during the 5 days is \(3 \mathrm{~kg}\). This corresponds to option (b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure and Temperature Relation
When it comes to gases, the relationship between pressure and temperature is crucial. In this exercise, the gas in the cylinder starts at a specific temperature and pressure. Pressure is a measure of the force exerted by gas particles when they collide with the walls of the container, while temperature measures the average kinetic energy of those particles. As the temperature remains constant in this scenario, according to the ideal gas law, pressure changes will depend on the amount of gas present. The Ideal Gas Law formula, \[PV = nRT\] explains that pressure \(P\), volume \(V\), and the number of moles \(n\) of a gas are proportionally related when the temperature \(T\) and the ideal gas constant \(R\) are fixed. So, if the temperature in Kelvin stays the same and the amount of gas changes, the pressure will vary accordingly.Key points to remember:
  • If the amount of gas decreases, pressure decreases provided temperature and volume remain constant.
  • Since temperature is constant in this scenario, any pressure change is related to a change in gas amount.
Molar Mass Calculations
Molar mass represents the mass of a substance's molecules in grams per mole. For the calculations in this problem, understanding molar mass is crucial to finding out how much gas remains or has leaked.For butane, the molar mass is given as \(58.12 \, \text{g/mol}\). With this information, calculating the number of moles present initially in 15 kg of gas is straightforward. Convert kilograms to grams first, then divide by molar mass:\[n_1 = \frac{15000}{58.12}\]The calculations here show how many moles of butane were initially in the container. Understanding these conversions is very important in chemistry because it allows us to switch between mass and moles easily. This step is essential for plugging into the Ideal Gas Law and understanding how much gas remains after some has leaked.
Gas Leaks and Pressure Changes
Gas leaks can lead to significant pressure changes as demonstrated in this problem. Initially, the pressure is measured at \(10 \, \text{atm}\) and falls to \(8 \, \text{atm}\) after leaking occurs without a temperature change.This scenario uses an important application of the Ideal Gas Law. With constant temperature and volume, changes in pressure illustrate that some of the gas has left the system. Calculating the final moles after the leak requires solving:\[n_2 = 0.8n_1\]Where \(n_2\) represents the remaining moles. This tells us 80% of the original gas remains, meaning 20% has leaked. Translating this percentage into mass:\[\text{Leaked mass} = 15 \, \text{kg} - 12 \, \text{kg} = 3 \, \text{kg}\]Key understanding from this example includes:
  • How pressure decrease indicates a loss of gas.
  • The relation between pressure, remaining moles, and how to calculate leaked gas.
  • Practical application in everyday issues, such as detecting leaks in gas cylinders.

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Most popular questions from this chapter

Relation between the three types of velocities, i.e., most probable velocity : average velocity : root mean square velocity is (a) \(\sqrt{3}: \sqrt{2}: \sqrt{\frac{8}{\pi}}\) (b) \(\sqrt{3}: \sqrt{2}: \sqrt{8}\) (c) \(\sqrt{2}: \sqrt{(8 / \pi)}: \sqrt{3}\) (d) \(1: 2: 3\)

Positive deviation from ideal behaviour takes place because of (a) molecular interaction between atoms and \(\frac{\mathrm{PV}}{\mathrm{nRT}}>1\) (b) molecular interaction between atoms and \(\frac{\mathrm{PV}}{\mathrm{nRT}}<1\) (c) finite size of atoms and \(\frac{\mathrm{PV}}{\mathrm{nRT}}>1\) (d) finite size of atoms and \(\frac{\mathrm{PV}}{\mathrm{nRT}}<1\)

In Haber's process, \(30 \mathrm{~L}\) of dihydrogen and \(30 \mathrm{~L}\) of dinitrogen were taken for reaction which yielded only \(50 \%\) of expected product. What is the composition of the gaseous mixture under afore-said conditions in the end? (a) \(20 \mathrm{~L} \mathrm{NH}_{3}, 25 \mathrm{~L} \mathrm{~N}_{2}, 15 \mathrm{~L} \mathrm{H}_{2}\) (b) \(20 \mathrm{~L} \mathrm{NH}_{3}, 20 \mathrm{~L} \mathrm{~N}_{2}, 20 \mathrm{~L} \mathrm{H}_{2}\) (c) \(10 \mathrm{~L} \mathrm{NH}_{3}, 25 \mathrm{~L} \mathrm{~N}_{2}, 15 \mathrm{~L} \mathrm{H}_{2}\) (d) \(20 \mathrm{~L} \mathrm{NH}_{3}, 10 \mathrm{~L} \mathrm{~N}_{2}, 30 \mathrm{~L} \mathrm{H}_{2}\)

A balloon having weight \(50 \mathrm{~kg}\) is filled with \(685.2 \mathrm{~kg}\) of helium gas at \(760 \mathrm{~mm}\) pressure and \(25^{\circ} \mathrm{C}\). What will be its pay load if it displaces \(5108 \mathrm{~kg}\) of air? (a) \(4372.8 \mathrm{~kg}\) (b) \(4392.6 \mathrm{~kg}\) (c) \(4444.4 \mathrm{~kg}\) (d) \(3482.9 \mathrm{~kg}\)

If a real gas follows equation \(\mathrm{P}(\mathrm{V}-\mathrm{nb})=\mathrm{RT}\) at low pressure, then for a graph between d/P vs. P, (where \(\mathrm{d}\) is the density of gas) (a) Intercept is \(\frac{\mathrm{MR}}{\mathrm{T}}\) (b) Intercept is \(\frac{\mathrm{M}}{\mathrm{RT}}\) (c) Slope is \(-\frac{b}{M(R T)^{2}}\) (d) Slope is \(-\frac{\mathrm{Mb}}{(\mathrm{RT})^{2}}\)
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