Problem 65 - University Physics with Modern Physics

Question

1 .00 mol of molecular nitrogen gas expands in volume very quickly, so no heat is exchanged with the environment during the process. If the volume increases from \(1.00 \mathrm{~L}\) to \(1.50 \mathrm{~L},\) determine the work done on the environment if the gas's temperature dropped from \(22.0^{\circ} \mathrm{C}\) to \(18.0^{\circ} \mathrm{C}\). Assume ideal gas behavior.

Step-by-Step Solution

Answer

Answer: The work done on the environment during this process is 83.20 J.

Step 1: Convert Temperatures

First, convert the initial and final temperatures from Celsius to Kelvin: Initial temperature: \(T_1 = 22.0^{\circ}C + 273.15 = 295.15 K\) Final temperature: \(T_2 = 18.0^{\circ}C + 273.15 = 291.15 K\)

Step 2: Find Initial and Final Pressure

From the data given, we know the initial and final volumes and temperatures. We can use the Ideal Gas Law to find the initial and final pressures of the gas. Recall the Ideal Gas Law: \(PV = nRT\) Initial pressure: \(P_1 = \frac{nRT_1}{V_1} = \frac{(1.00\mathrm{~mol})(8.314\mathrm{~J/mol\cdot K})(295.15\mathrm{~K})}{1.00\mathrm{~L}}\cdot\frac{1000\mathrm{~L}}{1\mathrm{~m^3}} = 24469.23\mathrm{~Pa}\) Final pressure: \(P_2 = \frac{nRT_2}{V_2} = \frac{(1.00\mathrm{~mol})(8.314\mathrm{~J/mol\cdot K})(291.15\mathrm{~K})}{1.50\mathrm{~L}}\cdot\frac{1000\mathrm{~L}}{1\mathrm{~m^3}} = 16140.75\mathrm{~Pa}\)

Step 3: Find the Work Done on the Environment

Now we will use the first law of thermodynamics, which states that the change in internal energy of the system is equal to the heat added to the system minus the work done by the system, or \(\Delta U = Q - W\). Since no heat is exchanged with the environment during the process (Q = 0), we can rewrite the equation as \(\Delta U = -W\). For an ideal gas, the change in internal energy is directly related to the change in temperature and can be represented by the equation \(\Delta U = nC_v\Delta T\). Here, \(C_v\) is the molar heat capacity at constant volume for molecular nitrogen, which is approximately 20.8 J/mol∙K. Calculate the change in internal energy: \(\Delta U = nC_v\Delta T = (1.00\mathrm{~mol})(20.8\mathrm{~J/mol\cdot K})(291.15\mathrm{~K} - 295.15\mathrm{~K}) = -83.20\mathrm{~J}\) Now, using the first law of thermodynamics equation, we can determine the work done on the environment: \(W = -\Delta U = -(-83.20\mathrm{~J}) = 83.20\mathrm{~J}\) So, the work done on the environment during this quick expansion process is 83.20 J.

Key Concepts

Work and Energy in Thermodynamics

In thermodynamics, understanding how work and energy interact is crucial, especially when no heat is exchanged with the environment. This situation is common in an adiabatic process, where the system is insulated. Here, the work done by or on the gas depends directly on the change in internal energy.

The first law of thermodynamics is key in these analyses. It 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 the work done by the system.

• If no heat is exchanged (\( Q = 0 \)), it simplifies to \( \Delta U = -W \).

This means the work done by or on the system directly correlates with the change in internal energy.

In this exercise, the internal energy change was calculated from the molar heat capacity and the change in temperature. The work done on the environment was equal to the negative of this change, indicating a strong interplay between work and energy in the absence of heat transfer.

Molar Heat Capacity

Molar heat capacity is an important concept when calculating changes in internal energy. It represents the amount of heat required to change the temperature of one mole of a substance by one degree Kelvin.

• For an ideal gas, the molar heat capacity can be at constant volume \(C_v\) or constant pressure \(C_p\).

In this scenario, since the volume is constant during part of the process, we utilize \(C_v\). For molecular nitrogen, \(C_v\) is about 20.8 J/mol\(\cdot\)K.

The relationship between changes in temperature and internal energy is expressed as:

• \( \Delta U = nC_v\Delta T \)

where:

• \(n\) is the number of moles,
• \(\Delta T\) is the change in temperature.

This formula helps us predict how the internal energy of the system changes when heated or cooled under constant volume conditions. In cases like this, where the temperature decreases, \(\Delta U\) will be negative, reflecting energy leaving the system as work.

Adiabatic Process

An adiabatic process is characterized by the absence of heat exchange with the surroundings, making it a fundamental concept in certain thermodynamic transformations. In these processes:

• No heat enters or leaves the system (\(Q = 0\)).
• Changes in pressure, volume, and temperature are all governed internally.

Because no heat is transferred, any change in temperature directly results in a change in internal energy. The ideal gas law \(PV = nRT\) can help describe such transformations, but it requires adjustment for adiabatic conditions.

In an adiabatic transformation, the form of the process can be visualized with relationships like:

• \(PV^\gamma = ext{constant} \)

Here, \(\gamma\) is the heat capacity ratio \(C_p/C_v\).

The given exercise confirms these principles as the molecular nitrogen gas expands without exchanging heat, seeking internal equilibrium. This quick expansion results in work done purely by the shift in internal energy, reflected in the computed work measure of 83.20 J done on the surrounding environment.