Problem 41 - University Physics with Modern Physics

Question

(a) How much heat does it take to increase the temperature of 2.50 mol of a diatomic ideal gas by 50.0 \(\mathrm{K}\) near room temperature if the gas is held at constant volume? (b) What is the answer to the question in part (a) if the gas is monatomic rather than diatomic?

Step-by-Step Solution

Answer

(a) 2598.1 J for diatomic gas; (b) 1558.9 J for monatomic gas.

Step 1: Understand the Problem and Given Data

We are tasked with finding the heat required to change the temperature of a given amount of gas. For (a) the gas is diatomic, and for (b) it's monatomic, with both conditions at constant volume. Given information includes 2.50 mol of gas and a temperature change of 50.0 K.

Step 2: Recall Heat Capacity for Ideal Gases

For a constant volume process, the heat capacity, denoted as \(C_v\), determines the heat added to the system. For diatomic gases \(C_v = \frac{5}{2}R\), and for monatomic gases \(C_v = \frac{3}{2}R\), where \(R\) is the universal gas constant, \(R = 8.314 \text{ J/mol.K}\).

Step 3: Calculate Heat for Diatomic Gas

Using the formula for heat at constant volume, \(Q = nC_v \Delta T\), substitute \(n = 2.50 \text{ mol}\), \(C_v = \frac{5}{2} \times 8.314 \text{ J/mol.K}\), and \(\Delta T = 50.0 \text{ K}\). Calculate \[ Q = 2.50 \times \frac{5}{2} \times 8.314 \times 50.0 \].

Step 4: Evaluate the Heat for Diatomic Gas

Perform the calculation from the previous step: \(Q = 2.50 \times 20.785 \times 50.0 = 2598.125 \text{ J}\). Thus, the heat required is 2598.125 J.

Step 5: Calculate Heat for Monatomic Gas

Substitute the values into the formula \(Q = nC_v \Delta T\) for monatomic gases: \(C_v = \frac{3}{2} \times 8.314 \text{ J/mol.K}\). Calculate \[ Q = 2.50 \times \frac{3}{2} \times 8.314 \times 50.0 \].

Step 6: Evaluate the Heat for Monatomic Gas

Complete the calculation: \(Q = 2.50 \times 12.471 \times 50.0 = 1558.875 \text{ J}\). Therefore, the heat required is 1558.875 J.

Key Concepts

Ideal Gas

An ideal gas is a theoretical concept used in physics and chemistry to describe gases that perfectly follow the gas laws. The assumptions behind an ideal gas include no intermolecular forces between the particles and that particles occupy no volume. This simplifies calculations and allows scientists to use basic formulas to estimate the behavior of gases under different conditions.

Key properties of an ideal gas are captured by the ideal gas law:

• \[PV = nRT\]

where:

• \(P\) is the pressure
• \(V\) is the volume
• \(n\) is the number of moles
• \(R\) is the ideal gas constant
• \(T\) is the temperature in Kelvin

Despite the model's simplicity, it provides a good approximation for real gases under many conditions, particularly high temperature and low pressure.

Diatomic Gas

Diatomic gases are composed of molecules with two atoms. Common examples are hydrogen \((H_2)\), nitrogen \((N_2)\), and oxygen \((O_2)\). These gases possess rotational energy in addition to translational energy due to the presence of two atoms.
For diatomic gases, the heat capacity at constant volume \(C_v\) is higher than for monatomic gases because of this additional rotational degree of freedom. The formula for \(C_v\) for a diatomic ideal gas is given by:

• \[C_v = \frac{5}{2} R\]

where \(R\) is the universal gas constant.

This means it takes relatively more heat to change the temperature of a diatomic gas since energy must also be distributed into the rotational motions of the molecules. Consequently, the higher heat capacity affects calculations for heat added to a system under constant volume, as evidenced by our exercises.

Monatomic Gas

Monatomic gases consist of individual atoms rather than molecules made up of multiple atoms. Examples of such gases are helium \((He)\), neon \((Ne)\), and argon \((Ar)\). These gases typically exhibit simpler behavior because they have only translational energy.For monatomic gases in an ideal state, the heat capacity at constant volume \(C_v\) is:

• \[C_v = \frac{3}{2} R\]

where \(R\) is the universal gas constant. This lower \(C_v\) indicates that less energy is needed to change their temperature because there are fewer ways to store energy (i.e., only translational modes are active).
Thus, in scenarios involving temperature change calculations, monatomic gases require less heat compared to diatomic gases, under the same conditions. This distinction is crucial for understanding how different gases behave under thermal processes.