Problem 24 - University Physics with Modern Physics

Question

Three moles of an ideal monatomic gas expands at a constant pressure of 2.50 atm; the volume of the gas changes from \(3.20 \times 10^{-2} \mathrm{m}^{3}\) to \(4.50 \times 10^{-2} \mathrm{m}^{3} .\) (a) Calculate the initial and final temperatures of the gas. (b) Calculate the amount of work the gas does in expanding. (c) Calculate the amount of heat added to the gas. (d) Calculate the change in internal energy of the gas.

Step-by-Step Solution

Answer

Initial temperature: 325.03 K, Final temperature: 456.33 K, Work done: 3293.06 J, Heat added: 8201.69 J, Change in internal energy: 4908.63 J.

Step 1: Convert Pressure to SI Unit

The given pressure is in atmospheres (atm). We'll convert it to Pascals (Pa) since SI units should be used in calculations. 1 atm = 101,325 Pa. Therefore:\[ P = 2.50 \text{ atm} \times 101,325 \text{ Pa/atm} = 253,312.5 \text{ Pa} \]

Step 2: Calculate Initial Temperature

Using the ideal gas law, \( PV = nRT \), we calculate the initial temperature \( T_i \). We have:\[ P = 253,312.5 \text{ Pa}, \; V_i = 3.20 \times 10^{-2} \text{ m}^3, \; n = 3 \text{ moles}, \; R = 8.314 \text{ J/(mol K)} \]Rearranging for \( T_i \):\[ T_i = \frac{PV_i}{nR} = \frac{253,312.5 \times 3.20 \times 10^{-2}}{3 \times 8.314} \approx 325.03 \text{ K} \]

Step 3: Calculate Final Temperature

Similarly, using the final volume \( V_f \), we calculate the final temperature \( T_f \):\[ V_f = 4.50 \times 10^{-2} \text{ m}^3 \]\[ T_f = \frac{PV_f}{nR} = \frac{253,312.5 \times 4.50 \times 10^{-2}}{3 \times 8.314} \approx 456.33 \text{ K} \]

Step 4: Calculate Work Done

The work done by the gas during the expansion at constant pressure is given by \( W = P \Delta V \), where \( \Delta V = V_f - V_i \).\[ \Delta V = 4.50 \times 10^{-2} - 3.20 \times 10^{-2} = 1.30 \times 10^{-2} \text{ m}^3 \]\[ W = 253,312.5 \times 1.30 \times 10^{-2} \approx 3293.06 \text{ J} \]

Step 5: Calculate Heat Added

For a constant pressure process, the heat added \( Q \) is given by \( Q = nC_p \Delta T \), where \( C_p = \frac{5}{2} R \) for monatomic gas:\[ \Delta T = T_f - T_i = 456.33 - 325.03 = 131.30 \text{ K} \]\[ C_p = \frac{5}{2} \times 8.314 = 20.785 \text{ J/(mol K)} \]\[ Q = 3 \times 20.785 \times 131.30 \approx 8201.69 \text{ J} \]

Step 6: Calculate Change in Internal Energy

The change in internal energy \( \Delta U \) can be found using \( \Delta U = nC_v \Delta T \), where \( C_v = \frac{3}{2} R \) for monatomic gas:\[ C_v = \frac{3}{2} \times 8.314 = 12.471 \text{ J/(mol K)} \]\[ \Delta U = 3 \times 12.471 \times 131.30 \approx 4908.63 \text{ J} \]

Step 7: Final Calculation and Verification

Check that \( Q = \Delta U + W \) to verify our calculations:\[ Q = 8201.69 \text{ J}, \; \Delta U = 4908.63 \text{ J}, \; W = 3293.06 \text{ J} \]Adding work and change in internal energy:\[ \Delta U + W = 4908.63 + 3293.06 = 8201.69 \text{ J} \]This verifies the consistency of our calculations, suggesting each step was correct.

Key Concepts

Internal Energy

Internal energy is a crucial concept in thermodynamics. It refers to the total energy contained within a system. For an ideal gas, this energy is the sum of the kinetic energies of the gas particles. When we talk about changes in internal energy, we consider the effect of temperature changes on this kinetic energy.
In this exercise, the internal energy change is calculated using the formula \( \Delta U = nC_v \Delta T \), where \( n \) is the number of moles, \( C_v \) is the specific heat at constant volume, and \( \Delta T \) is the temperature difference. It's important to know that the formula relies on specific heat, which measures how much heat is needed to raise the temperature of a mole by one degree.
The calculation of internal energy doesn't depend on pressure changes directly. Instead, it's tied to how gas particles speed up with temperature increases. For an ideal monatomic gas, \( C_v = \frac{3}{2} R \) emphasizes how internal energy is married to the temperature of the gas.

Heat Transfer

Heat transfer involves the exchange of energy between the system (in this case, the gas) and its surroundings due to a temperature difference. In thermodynamics, when we heat an ideal gas, we're essentially transferring energy into the system.
In this constant pressure scenario, the formula \( Q = nC_p \Delta T \) defines how much heat adds to the gas, where \( C_p = \frac{5}{2} R \) is the specific heat at constant pressure. This equation demonstrates the relation between heat added and temperature change at constant pressure.
Heat plays a significant role; the gas absorbs it and uses part of this energy to increase the internal energy and the rest to perform work on the surroundings. Understanding these elements helps distinguish heat from temperature, a common confusion.

Work Done

The concept of work done by a gas expands our understanding of thermodynamic processes. When a gas changes volume at constant pressure, it does work on its surroundings, which we can calculate with \( W = P \Delta V \). Here \( P \) is the constant pressure of the gas and \( \Delta V \) is the change in volume.
This formula tells us that the work is a product of the force (pressure) and the distance (volume change). Importantly, this work is positive in expansion because the gas is doing work on the environment by pushing against external pressure.
It's essential to grasp how work and energy interrelate. Work done by the gas decreases the energy available for raising the temperature, linking back to heat transfer where some energy goes into doing this work.

Temperature Change

Temperature change connects directly to energy changes in a system. For an ideal gas in a constant pressure and volume process, temperature changes can illustrate how energy moves.
By using the ideal gas law \( PV = nRT \), we can figure out initial and final temperatures. In this exercise, changes in volume at constant pressure demonstrate this connection: as the gas expanded, both its volume and temperature rose. This is because the increase in temperature results from the external conditions applied and the gas properties.
Temperature measures the average kinetic energy of particles, so changing the temperature will influence internal energy and vice versa. This shows why the formula \( T = \frac{PV}{nR} \) plays a crucial role in determining how much energy is in the system.