Problem 29 - University Physics with Modern Physics

Question

The temperature of 0.150 mol of an ideal gas is held stant at \(77.0^{\circ} \mathrm{C}\) while its volume is reduced to 25.0\(\%\) of its initial volume. The initial pressure of the gas is 1.25 \(\mathrm{atm}\) . (a) Determine the work done by the gas. (b) What is the change in its internal energy? (c) Does the gas exchange heat with its surroundings? If so, how much? Does the gas absorb or liberate heat?

Step-by-Step Solution

Answer

(a) Work done by the gas is 1094 J. (b) Change in internal energy is 0. (c) Gas absorbs 1094 J of heat.

Step 1: Convert Temperature to Kelvin

To apply the Ideal Gas Law, we first need to convert the temperature from Celsius to Kelvin. The conversion formula is \( T(K) = T(^{\circ}C) + 273.15 \). Thus, the temperature in Kelvin is \( 77.0 + 273.15 = 350.15 \ K \).

Step 2: Calculate Final Pressure Using Boyle's Law

Since the temperature is constant, we use Boyle's Law \( P_1V_1 = P_2V_2 \). Given \( V_2 = 0.25V_1 \), the final pressure \( P_2 = \frac{P_1V_1}{V_2} = \frac{1.25 \times V_1}{0.25V_1} = 5.0 \ \mathrm{atm} \).

Step 3: Calculate Work Done by the Gas

The work done by the gas under constant temperature (isothermal process) is given by \( W = -nRT \ln \left( \frac{V_2}{V_1} \right) \). Plugging in values: \( n = 0.150 \ \mathrm{mol}, T = 350.15 \ \mathrm{K}, R = 0.0821 \ \mathrm{L\ atm \ mol^{-1} \ K^{-1}} \) and \( V_2/V_1 = 0.25 \, \), we get: \[ W = -0.150 \times 0.0821 \times 350.15 \times \ln(0.25) = 10.8 \, \mathrm{L\ atm} \].

Step 4: Convert Work from L atm to J

Convert the work done to Joules using the conversion factor: \( 1 \ \mathrm{L\ atm} = 101.3 \ \mathrm{J} \). Therefore, \( W = 10.8 \times 101.3 = 1094 \ \mathrm{J} \).

Step 5: Determine Change in Internal Energy

For an ideal gas undergoing an isothermal process, the change in internal energy \( \Delta U \) is zero, because the internal energy of an ideal gas depends only on temperature, which remains constant.

Step 6: Calculate Heat Exchange Using First Law of Thermodynamics

The First Law of Thermodynamics is \( \Delta U = Q - W \). Since \( \Delta U = 0 \), we find \( Q = W \). Therefore, the gas exchanges 1094 J of heat with its surroundings. The gas absorbs this amount of heat to do the work.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation used to describe the behavior of an ideal gas. It relates four important properties of the gas: pressure (\( P \)), volume (\( V \)), temperature (\( T \)), and the amount of substance in moles (\( n \)). The equation is given by:\[PV = nRT\]where \( R \) is the ideal gas constant. For this law to hold, it is crucial that the gas behaves ideally, meaning the interactions between molecules are negligible, and the volume occupied by the gas molecules is much smaller than the volume of the container. An essential aspect of this exercise is making sure temperature is measured in Kelvin. This is necessary because Kelvin is the absolute temperature scale that leads to a direct proportional relationship with pressure and volume in this equation.

Boyle's Law

Boyle's Law focuses on the relationship between pressure and volume when temperature and the amount of gas remain constant. It states that the pressure of a gas is inversely proportional to its volume, i.e., as the volume decreases, the pressure increases, and vice versa. This relationship is expressed by the equation:\[P_1V_1 = P_2V_2\]In this exercise, the temperature is kept constant and corresponds to an isothermal process. The volume of the gas is reduced to 25% of its initial volume. As Boyle's Law dictates, this causes the pressure to increase, and here, it is specifically calculated based on the reduced volume fraction.

First Law of Thermodynamics

The First Law of Thermodynamics is a version of the law of energy conservation, tailored for thermodynamic systems. It states that energy cannot be created or destroyed in an isolated system. Mathematically, it is expressed as:\[\Delta U = Q - W\]where \( \Delta U \) is the change in internal energy, \( Q \) is the heat exchanged with the surroundings, and \( W \) is the work done by the system. In an isothermal process involving an ideal gas, the internal energy change is zero, leading the heat absorbed or released to equal the work done.

Internal Energy Change

Internal energy is a key concept when discussing thermodynamics. For an ideal gas, the internal energy depends solely on its temperature. This is due to the fact that ideal gases have particles that move independently without interactions. In the given exercise, the process is isothermal, meaning the temperature remains constant throughout. Consequently, the internal energy of the gas does not change. This means that any work done by the gas must be compensated by an equal amount of heat absorbed, thereby maintaining the energy balance in accordance with the First Law of Thermodynamics. The zero change in internal energy in an isothermal process is an insightful aspect that can simplify solving related thermodynamic problems.