Question

A sample of argon gas has a volume of \(735 \mathrm{~mL}\) at a pressure of \(1.20 \mathrm{~atm}\) and a temperature of \(112{ }^{\circ} \mathrm{C}\). What is the volume of the gas, in milliliters, when the pressure and temperature of the gas sample are changed to the following, if the amount of gas remains the same? a. \(658 \mathrm{mmHg}\) and \(281 \mathrm{~K}\) b. \(0.55 \mathrm{~atm}\) and \(75^{\circ} \mathrm{C}\) c. \(15.4 \mathrm{~atm}\) and \(-15^{\circ} \mathrm{C}\)

Short answer

a. \(760.77 \mathrm{~mL}\), b. \(1797.69 \mathrm{~mL}\), c. \(14.96 \mathrm{~mL}\)

Step-by-step solution

- Write down the Ideal Gas Law

The Ideal Gas Law is given by: \[ PV = nRT \] Where: - \(P\) is the pressure - \(V\) is the volume - \(n\) is the number of moles of the gas - \(R\) is the gas constant - \(T\) is the temperature

- Use Combined Gas Law

Since the amount of gas remains the same, use the Combined Gas Law: \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \] Rearrange it to solve for the new volume, \( V_2 \), \[ V_2 = \frac{P_1V_1T_2}{P_2T_1} \]

- Convert Initial Conditions to Correct Units

Initial volume, \(V_1\), is given as \(735 \mathrm{~mL}\). Initial pressure, \(P_1\), is given as \(1.20 \mathrm{~atm}\). Initial temperature, \(T_1\), is converted to Kelvin: \[ T_1 = 112 + 273 = 385 \mathrm{~K} \]

- Solve for Volume in Case a

Given: \(P_2 = 658 \mathrm{~mmHg}\) (convert to atm: \( P_2 = \frac{658}{760} \mathrm{~atm} \approx 0.866 \mathrm{~atm} \)) \(T_2 = 281 \mathrm{~K}\) Using the Combined Gas Law: \[ V_2 = \frac{1.20 \times 735 \times 281}{0.866 \times 385} \approx 760.77 \mathrm{~mL} \]

- Solve for Volume in Case b

Given: \(P_2 = 0.55 \mathrm{~atm}\) \(T_2 = 75^{\circ} \mathrm{C}\) (convert to Kelvin: \( T_2 = 75 + 273 = 348 \mathrm{~K} \)) Using the Combined Gas Law: \[ V_2 = \frac{1.20 \times 735 \times 348}{0.55 \times 385} \approx 1797.69 \mathrm{~mL} \]

- Solve for Volume in Case c

Given: \(P_2 = 15.4 \mathrm{~atm}\) \(T_2 = -15^{\circ} \mathrm{C}\) (convert to Kelvin: \( T_2 = -15 + 273 = 258 \mathrm{~K} \)) Using the Combined Gas Law: \[ V_2 = \frac{1.20 \times 735 \times 258}{15.4 \times 385} \approx 14.96 \mathrm{~mL} \]

Key concepts

Ideal Gas Law

The Ideal Gas Law is an essential principle in chemistry and physics that helps us understand how gases behave under different conditions. The law is represented by the equation: \( PV = nRT \) - \(P\) stands for pressure - \(V\) stands for volume - \(n\) stands for the number of moles of the gas - \(R\) is the gas constant - \(T\) stands for temperature in Kelvin This equation shows the relationship between the pressure, volume, temperature, and amount of gas. It assumes that the gas is ideal, meaning its particles do not interact and occupy no volume.

Volume Calculation

When we need to find the volume of a gas under changing conditions, we can use the Combined Gas Law. This is particularly helpful when the amount of gas remains the same. The formula is: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \) Here's how you can solve for the new volume (\( V_2 \)) when you know the initial and final conditions: \( V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \) This equation allows us to link the initial and final states of the gas by incorporating pressure, volume, and temperature to find the solution for volume easily.

Pressure Conversion

It's crucial to have all your pressure measurements in the same unit before using gas laws. Often, you might need to convert between different units of pressure, such as mmHg and atm. For instance, to convert from mmHg to atm, use the conversion factor: - \( 1 \text{ atm} = 760 \text{ mmHg} \) Given a pressure of 658 mmHg, you can convert it to atm as follows: \( P_2 = \frac{658}{760} \approx 0.866 \text{ atm} \) Having all pressures in atm (or any consistent unit) simplifies calculations and reduces the chance of errors.

Temperature Conversion

In gas laws, temperature must always be in Kelvin. Here's why and how you can convert common units to Kelvin. The Kelvin scale starts at absolute zero, which is essential for accurate gas law calculations because these laws assume absolute pressure and volume. The conversion from Celsius to Kelvin is straightforward: - \( T_K = T_{C} + 273 \) For example: converting 112°C to Kelvin: - \( T_1 = 112 + 273 = 385 \text{ K} \) Whether you are starting with a positive Celsius value (like 75°C) or a negative one (like -15°C), always add 273 to get the Kelvin temperature, necessary for accurate gas calculations.