Question

What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) The initial pressure is 722 torr, and the final pressure is \(0.950 \mathrm{~atm}\) the initial temperature is \(32^{\circ} \mathrm{F},\) and the final temperature is \(273 \mathrm{~K}\). (b) Half the gas escapes (at constant \(P\) and \(T\) ). (c) Both the pressure and temperature decrease to \(\frac{1}{4}\) of their initial values.

Short answer

(a) No change, (b) decreased by half, (c) no change.

Step-by-step solution

- Understand the Ideal Gas Law

The Ideal Gas Law is given by the equation \[ PV = nRT \] where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature.

- Convert Units

Convert the initial temperature from Fahrenheit to Kelvin. \[ T_{initial} (K) = \frac{5}{9} (32 \degree F - 32) + 273.15 = 273.15 K \]

- Scenario (a): Apply Ideal Gas Law

Initial state: \( P_{initial} = 722 \text{ torr} = 0.950 \text{ atm} \)\( T_{initial} = 273.15 \text{ K} \) Final State: \( P_{final} = 0.950 \text{ atm} \)\( T_{final} = 273 \text{ K} \)Use the law to find volume ratio: \[ V_{final} / V_{initial} = \frac{P_{initial} T_{final}}{P_{final} T_{initial}} = \frac{722 \cdot 273}{0.950 \cdot 273.15} = 1 \]

- Scenario (b): Apply Ideal Gas Law

Half the gas escapes, hence the number of moles is halved while pressure and temperature remain the same. \[ V_{final} = \frac{1}{2} V_{initial} \]

- Scenario (c): Apply Ideal Gas Law

Both the pressure and temperature decrease to \( \frac{1}{4} \) of their initial values. Use the law to find the volume ratio:\[ V_{final} / V_{initial} = \frac{P_{initial} T_{final}}{P_{final} T_{initial}} = \frac{P_{initial}} {\frac{1}{4} P_{initial}} \cdot \frac{\frac{1}{4} T_{initial}}{T_{initial}} = \frac{4}{1} \cdot \frac{1}{4} = 1 \]

Key concepts

Pressure-Volume Relationship

The pressure-volume relationship in gases is one of the cornerstones of understanding how gases behave. According to the Ideal Gas Law, this relationship is described by the equation: \(PV = nRT\).Here, \(P\) represents the pressure of the gas, \(V\) is the volume, \(n\) denotes the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

When you change the pressure of gas while keeping the temperature and number of moles constant, the volume of the gas will adjust to maintain the relationship in the equation. Specifically, volume and pressure have an inverse relationship when temperature is constant.

• As pressure increases, volume decreases.
• As pressure decreases, volume increases.

In scenario (a) from the exercise, we see that even when the final pressure and temperature both change, the product \(PV\) must stay constant if the number of moles and the gas constant \(R\) do not change. Thus, carefully balancing the changes helps us calculate the final volume of gas.

Temperature Conversion

Another key element to solving gas law problems is accurately converting temperatures to Kelvin, as the Ideal Gas Law relies on absolute temperature. Given temperatures in Fahrenheit or Celsius must be converted for precise calculations.

The formula to convert Fahrenheit to Kelvin is: \(T_{K} = \frac{5}{9} (T_{F} - 32) + 273.15\). In the provided step-by-step solution, converting the initial temperature of \(32^{\bit F}\) to Kelvin involves applying this formula: \(T_{initial} = \frac{5}{9}(32^{\bit F} - 32) + 273.15 = 273.15 \text{ K}\).

It's important not to skip the temperature conversion step, as incorrect temperatures would lead to erroneous volume calculations. Consistency in units ensures that you can apply the Ideal Gas Law correctly, leading to meaningful results.

Remember, always convert to Kelvin in gas law problems to maintain accuracy!

Number of Moles Effect

The number of moles of gas \(n\) directly affects the volume. This relationship is crucial in applying the Ideal Gas Law. If you change the number of moles while keeping pressure and temperature constant, the volume will change proportionally.

In scenario (b), half of the gas escapes, which means the number of moles is halved. This leads to a proportional decrease in volume since the pressure \(P\) and temperature \(T\) remain the same. Thus, the new volume would be:\(V_{final} = \frac{1}{2} V_{initial}\).

This principle allows us to understand how changes in the amount of gas affect the space it occupies under constant conditions. Always note how adjusting the number of moles (by adding or removing gas) will impact volume using:

• More moles of gas occupy more volume.
• Fewer moles result in a smaller volume.

Changes in \(n\) are especially useful in chemical reactions where gases are produced or consumed, impacting the resulting gas volumes in controlled environments.