Problem 52 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The partial pressures of \(\mathrm{N}_{2}, \mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) in a vessel are \(38 \mathrm{~cm}\) of \(\mathrm{Hg}, 190\) torr and \(0.5\) atm, respectively. The total pressure of the mixture at the same temperature is (a) \(0.96 \mathrm{~atm}\) (b) \(1.02 \mathrm{~atm}\) (c) \(1.64\) atm (d) \(1.25 \mathrm{~atm}\)

Step-by-Step Solution

Answer

Correct answer is (d) 1.25 atm, after converting all partial pressures to atm and summing them up.

Step 1 - Convert all pressures to the same unit

First, we need to convert the partial pressures of the gases into the same unit. We can convert cm of Hg and torr to atm because 1 atm = 760 torr and 1 torr = 1 mm of Hg = 1 cm of Hg / 10. Therefore, the partial pressure of \(\mathrm{N}_2\) is \(38 \text{ cm of Hg} \times \frac{1 \text{ torr}}{1 \text{ cm of Hg}} \times \frac{1 \text{ atm}}{760 \text{ torr}}\) = \(\frac{38}{760} \text{ atm}\) and for \(\mathrm{O}_2\) is \(190 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}}\) = \(\frac{190}{760} \text{ atm}\). The partial pressure of \(\mathrm{CO}_2\) is already in atm.

Step 2 - Calculate the pressures in atm

Calculate each gas's partial pressure in atm. For \(\mathrm{N}_2\), it is \(\frac{38}{760} \text{ atm} \) or approximately 0.05 atm. For \(\mathrm{O}_2\), it is \(\frac{190}{760} \text{ atm} \) or approximately 0.25 atm. The partial pressure of \(\mathrm{CO}_2\) remains 0.5 atm.

Step 3 - Add the partial pressures

Add the partial pressures of all three gases to find the total pressure. Total pressure \(P_{\text{total}} = P_{\mathrm{N}_2} + P_{\mathrm{O}_2} + P_{\mathrm{CO}_2} = 0.05 \text{ atm} + 0.25 \text{ atm} + 0.5 \text{ atm} = 0.8 \text{ atm}\).

Step 4 - Determine the correct answer

The total calculated pressure (0.8 atm) is not listed as an option, indicating a possible calculation error. Re-evaluate the calculation.

Step 5 - Correcting the calculation

Recalculate the partial pressures and add them accurately. For \(\mathrm{N}_2\): \(\frac{38}{760} \text{ atm} = 0.05 \text{ atm}\), for \(\mathrm{O}_2\): \(\frac{190}{760} \text{ atm} = 0.25 \text{ atm}\), but \(\mathrm{CO}_2\) was already provided in atm: 0.5 atm. So, \(P_{\text{total}} = 0.05 \text{ atm} + 0.25 \text{ atm} + 0.5 \text{ atm} = 0.8 \text{ atm}\). However, there was a miscalculation in converting \(\mathrm{N}_2\)'s partial pressure. It should be \(\frac{38}{760} \text{ atm}\) = 0.05 atm. Correct calculations give \(P_{\text{total}} = 0.05 \text{ atm} + 0.25 \text{ atm} + 0.5 \text{ atm} = 0.8 \text{ atm}\), which is still inconsistent with the given options. It seems there has been a mistake in the calculations or the options provided are incorrect.

Step 6 - Recheck conversion factors and redo calculations

Upon careful reevaluation, the conversion of cm of Hg to atm is done by dividing by 76 (and not 760) because 1 atm = 76 cm of Hg. Now, correct the pressure of \(\mathrm{N}_2\) to: \(38 \text{ cm of Hg} \times \frac{1 \text{ atm}}{76 \text{ cm of Hg}}\) = 0.5 atm. Correct the pressure for \(\mathrm{O}_2\) using torr as: \(190 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}}\) = 0.25 atm. Adding these up with \(\mathrm{CO}_2\)'s pressure gives: \(0.5 + 0.25 + 0.5 \text{ atm} = 1.25 \text{ atm}\) which matches option (d).

Key Concepts

Partial Pressure

Partial pressure is a key concept in understanding gas mixtures. It's the pressure that an individual gas in a mixture would exert if it were alone in the container.

Imagine a group of different gases living together in a room; each gas behaves independently, not knowing about the others around it. The partial pressure is like the personal space that gas claims in the room. If you combine all the personal spaces (partial pressures), you get the total pressure, which is like everyone standing together shoulder to shoulder.

• Represented as: \( P_{gas} \)
• To find total pressure, sum up each gas's partial pressure.
• Important for applications like calculating the concentration of oxygen in breathable air.

Using this concept, we ensure a better understanding of the exercise at hand, watch out for possible miscalculations that can occur when hastily summing up partial pressures.

Pressure Conversion

Pressure conversion is all about speaking the same language. There are multiple units for measuring pressure – such as atm, torr, mmHg, and cmHg. To accurately calculate total pressure or use the ideal gas law, we need to convert all pressures into the same unit.

The conversion factor connects different pressure languages and allows us to translate one into another, ensuring we can accurately combine pressures without making errors.

• Common conversions: 1 atm = 76 cmHg = 760 mmHg = 760 torr.
• Always check your conversion factors to avoid mistakes

In exercises dealing with pressures in various units, precision in converting these values is crucial to arriving at the correct answer.

Ideal Gas Equation

The ideal gas equation (\(PV=nRT\)) is a superstar in the world of gases, capturing their behavior with elegance. It shows how pressure (\(P\)), volume (\(V\)), number of moles (\(n\)), temperature (\(T\)), and the gas constant (\(R\)) are all related.

If a gas perfectly followed this equation, it would be called an 'ideal gas.' Real gases don't always play by these rules, especially under extreme conditions, but for many situations, it's a reliable model.

• Helps predict the behavior of gases when we change conditions like pressure or temperature.
• Crucial in chemical reactions involving gases.

Understanding the ideal gas equation allows students to engage with problems confidently, manipulate variables, and comprehend the interdependence of these variables.

Mole Concept

The mole concept is a gateway to understanding chemical quantities. A 'mole' is simply a huge number of particles, exactly like a 'dozen' eggs means 12 eggs; a mole means \(6.022 \times 10^{23}\) particles.

This concept allows chemists to count atoms or molecules by weighing them and connects the microscopic world of atoms to the macroscopic world we can measure.

• Central to converting between mass, volume, and particle count in chemistry.
• Directly connected with gas laws, allowing quantity calculation from given pressures and volumes.

Incorporating the mole concept seamlessly into gas law problems ensures accurate calculation of gas quantities and deepens understanding of chemical reactions.