Problem 27 - University Physics with Modern Physics

Question

A cylinder with a movable piston contains 3.00 \(\mathrm{mol}\) of \(\mathrm{N}_{2}\) gas (assumed to behave like an ideal gas). (a) The \(\mathrm{N}_{2}\) is heated at constant volume until 1557 \(\mathrm{J}\) of heat have been added. Calculate the change in temperature. (b) Suppose the same amount of heat is added to the \(\mathrm{N}_{2}\) , but this time the gas is allowed to expand while remaining at constant pressure. Calculate the temperature change. (c) In which case, (a) or \((b),\) is the final internal energy of the \(\mathbf{N}_{2}\) higher? How do you know? What accounts for the difference between the two cases?

Step-by-Step Solution

Answer

(a) \( \Delta T \approx 25.0 \, \text{K} \); (b) \( \Delta T \approx 17.9 \, \text{K} \); (a) shows higher internal energy.

Step 1: Identify the Problem Type

This problem involves the first law of thermodynamics in which an ideal gas receives heat under different constraints (constant volume and constant pressure). We'll use specific heat capacities for each process.

Step 2: Calculate Temperature Change for Constant Volume Process (Part a)

For a constant volume process, use the formula for the change in internal energy: \[ \Delta U = Q = nC_v \Delta T \] where \( Q = 1557 \, \text{J} \), \( n = 3.00 \, \text{mol} \), and \( C_v = \frac{5}{2}R \). Solving for \( \Delta T \):\[ \Delta T = \frac{Q}{nC_v} = \frac{1557}{3.00 \cdot \frac{5}{2} \cdot 8.31} \approx 25.0 \, \text{K} \]

Step 3: Calculate Temperature Change for Constant Pressure Process (Part b)

For a constant pressure process, use the formula for heat added: \[ Q = nC_p \Delta T \] where \( Q = 1557 \, \text{J} \), \( n = 3.00 \, \text{mol} \), and \( C_p = \frac{7}{2}R \). Solving for \( \Delta T \):\[ \Delta T = \frac{1557}{3.00 \cdot \frac{7}{2} \cdot 8.31} \approx 17.9 \, \text{K} \]

Step 4: Compare Final Internal Energies (Part c)

The internal energy change is higher in the constant volume process. This is because all the heat added goes into increasing internal energy when volume is constant, whereas in the constant pressure process, some energy is used to perform work as the gas expands. Therefore, in case (a), the final internal energy of the \(\mathrm{N}_2\) is higher.

Key Concepts

Ideal Gas

Understanding the concept of an ideal gas is fundamental when analyzing processes like heating and expansion. An ideal gas is a theoretical gas composed of many randomly moving point particles that do not interact except when they collide elastically. This means that:

• It follows the ideal gas law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.
• Intermolecular forces and the volume of individual gas particles are neglected.

This helps simplify calculations, especially during thermodynamic processes. In the exercise, the nitrogen gas \( N_2 \) is an ideal gas, allowing us to use simple relationships for constant volume and constant pressure processes. By understanding these assumptions, you can better appreciate how energy, in the form of heat, affects temperature changes and internal energy.

Constant Volume Process

In a constant volume process, the volume of the gas remains fixed even though it absorbs heat. This is a special case where the system does not perform work, as work is defined as "force times distance," and without volume change, there is no distance.The First Law of Thermodynamics states:\[ \Delta U = Q - W \]where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is work done by the system. In a constant volume process, since \( W = 0 \), the equation simplifies to:\[ \Delta U = Q \]This means all the heat energy \( Q \) supplied goes into changing the internal energy \( \Delta U \). In the given exercise, applying 1557 J of heat directly increases the temperature of the nitrogen gas by about 25.0 K. This process showcases how heat input directly affects a gas's internal energy when volume is kept constant.

Constant Pressure Process

In a constant pressure process, the pressure of the gas is maintained while allowing the volume to change. Here, the gas does perform work on its surroundings, as it can expand or contract as it absorbs or releases heat.The thermodynamic equation for such a process is:\[ Q = \Delta U + W \]Work done by the gas at constant pressure is given by:\[ W = P \Delta V \]This means that some heat energy gets converted to work, typically related to volume change, leaving less energy to increase the internal temperature compared to a constant volume process. As a result, the temperature change is slightly smaller.In the exercise, when 1557 J of heat is added, the temperature changes by about 17.9 K. This highlights the trade-off between increasing internal energy and performing work during expansion, underlining the complexity of real-world applications where different constraints apply.