Problem 87
Question
A \(2.24\) L cylinder of oxygen at NTP is found to develop a leakage. When the leakage was plugged the pressure dropped to \(570 \mathrm{~mm}\) of \(\mathrm{Hg}\). The number of moles of gas that escaped will be (a) \(0.050\) (b) \(0.025\) (c) \(0.075\) (d) \(0.01\)
Step-by-Step Solution
Verified Answer
(b) 0.025
1Step 1: Understanding the Problem
We need to find the number of moles of oxygen gas that escaped from the cylinder. Initially, the gas is at NTP (Normal Temperature and Pressure). After plugging a leakage, the pressure drops to 570 mm Hg. We will use the ideal gas law and the initial and final conditions to solve this problem.
2Step 2: Initial Conditions at NTP
NTP conditions mean that the pressure is 1 atm (or 760 mm Hg), the temperature is 273.15 K, and the volume is 2.24 L. We will use the ideal gas law, \( PV = nRT \), to find the initial number of moles in the gas cylinder.
3Step 3: Calculate Initial Moles
Using the ideal gas law \( PV = nRT \) with initial conditions: \( P = 760 \) mm Hg, \( V = 2.24 \) L, \( R = 62.36 \) L mm Hg/mol K (appropriate R for these units), and \( T = 273.15 \) K, calculate the initial moles \( n_1 \).\[ n_1 = \frac{PV}{RT} = \frac{760 \times 2.24}{62.36 \times 273.15} \approx 0.1 \text{ mol} \]
4Step 4: Final Conditions After Leakage
After plugging the leakage, the pressure drops to 570 mm Hg while temperature and volume remain the same. Calculate the final number of moles \( n_2 \) using the ideal gas law for these new conditions.
5Step 5: Calculate Final Moles
Use \( PV = nRT \) with \( P = 570 \) mm Hg and the same \( V, R, T \) to find \( n_2 \).\[ n_2 = \frac{570 \times 2.24}{62.36 \times 273.15} \approx 0.075 \text{ mol} \]
6Step 6: Determine Moles Escaped
Calculate the difference between the initial and final moles to find the amount of gas that escaped. \[ \text{Moles Escaped} = n_1 - n_2 = 0.1 - 0.075 = 0.025 \text{ mol} \]
Key Concepts
Normal Temperature and Pressure (NTP)Moles of gasPressure change in gases
Normal Temperature and Pressure (NTP)
Normal Temperature and Pressure (NTP) is a set of conditions many scientists use as a reference point for measuring gases. Under NTP:
- Pressure is considered to be 1 atmosphere (atm), which is equivalent to 760 mm Hg.
- Temperature is set at 273.15 Kelvin (K), the freezing point of water under standard atmospheric conditions.
- This standardization helps in making consistent and comparable calculations involving gases.
Moles of gas
The concept of moles is essential when working with gases, particularly with the ideal gas law, which is expressed as \( PV = nRT \). Here,
In the problem, we initially calculated the moles of gas at NTP and then again after some gas had escaped. By applying the initial conditions into the equation, we found the initial moles of gas. With the same method, adjusting for the pressure change, we calculated how many moles remained. This step is crucial to find out how much gas escaped the cylinder.
- \( P \) is the pressure of the gas,
- \( V \) is the volume,
- \( n \) represents the number of moles,
- \( R \) is the ideal gas constant that depends on the units used,
- \( T \) is the temperature in Kelvin.
In the problem, we initially calculated the moles of gas at NTP and then again after some gas had escaped. By applying the initial conditions into the equation, we found the initial moles of gas. With the same method, adjusting for the pressure change, we calculated how many moles remained. This step is crucial to find out how much gas escaped the cylinder.
Pressure change in gases
Pressure plays a significant role in gas behavior, especially when a leak in a system occurs. According to the ideal gas law, pressure, volume, and temperature can influence the number of moles present. In our problem, the pressure dropped from 760 mm Hg to 570 mm Hg due to a leakage:
In essence, the pressure reduction signifies a loss of moles or gas molecules within the system. Calculating this allows us to determine how much escaped, aiding in control and maintenance of gas systems.
- Initially, the gas was under the standard pressure of 1 atm (760 mm Hg) at NTP.
- After fixing the leak, the pressure dropped, while temperature and volume remained unchanged.
In essence, the pressure reduction signifies a loss of moles or gas molecules within the system. Calculating this allows us to determine how much escaped, aiding in control and maintenance of gas systems.
Other exercises in this chapter
Problem 84
Equal masses of methane and hydrogen are mixed in an empty container at \(25^{\circ} \mathrm{C}\). The fraction of the total pressure exerted by hydrogen is (a)
View solution Problem 85
Helium atom is two times heavier than a hydrogen molecule. At \(298 \mathrm{~K}\), the average kinetic energy of a helium atom is (a) same as that of a hydrogen
View solution Problem 89
A balloon having weight \(50 \mathrm{~kg}\) is filled with \(685.2 \mathrm{~kg}\) of helium gas at \(760 \mathrm{~mm}\) pressure and \(25^{\circ} \mathrm{C}\).
View solution Problem 90
For non-zero value of force of attraction between gas molecules, gas equation will be (a) \(\mathrm{PV}=\mathrm{n} \mathrm{RT}-\frac{\mathrm{n}^{2}}{\mathrm{~V}
View solution