Problem 36 - University Physics with Modern Physics

Question

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at a pressure of \(1.00 \times 10^{5} \mathrm{Pa}\) and occupies a volume of \(2.50 \times 10^{-3} \mathrm{m}^{3}\) (a) Find the initial temperature of the gas in kelvins. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; ( (ii) isobaric; (iii) adiabatic.

Step-by-Step Solution

Answer

Initial temp: 300 K. Isothermal: 300 K, 5.00x10^4 Pa. Isobaric: 600 K, 1.00x10^5 Pa. Adiabatic: 238 K, 3.71x10^4 Pa.

Step 1: Use Ideal Gas Law to Find Initial Temperature

The ideal gas law is given by the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant (8.314 J/mol·K), and \( T \) is temperature in kelvins. Substitute the given values: \( P = 1.00 \times 10^{5} \) Pa, \( V = 2.50 \times 10^{-3} \) m³, and \( n = 0.100 \) mol.\[ T = \frac{PV}{nR} = \frac{(1.00 \times 10^{5})(2.50 \times 10^{-3})}{0.100 \times 8.314} \approx 300 \text{ K} \]

Step 2: Isothermal Expansion

During an isothermal process, the temperature remains constant. Since the initial temperature is 300 K, the final temperature for isothermal expansion is also 300 K. Using \( PV = \text{constant} \), we find the final pressure.\[ P_f V_f = P_i V_i \Rightarrow P_f (2V_i) = P_i V_i \Rightarrow P_f = \frac{P_i}{2} \]Thus, \( P_f = \frac{1.00 \times 10^{5}}{2} = 5.00 \times 10^{4} \) Pa.

Step 3: Isobaric Expansion

For an isobaric process, the pressure remains constant. Therefore, the final pressure is \( P_f = 1.00 \times 10^5 \) Pa. The volume doubles, and we use the ideal gas law to find the final temperature.\[ T_f = \frac{P_f V_f}{nR} = \frac{1.00 \times 10^{5} \times 5.00 \times 10^{-3}}{0.100 \times 8.314} \approx 600 \text{ K} \]

Step 4: Adiabatic Expansion

For adiabatic processes, we use the relation \( TV^{\gamma-1} = \text{constant} \) where \( \gamma = \frac{5}{3} \) for a monatomic ideal gas. We find the final temperature by:\[ T_f V_f^{\gamma-1} = T_i V_i^{\gamma-1} \]\[ T_f (2V_i)^{(\frac{5}{3} - 1)} = 300 \times V_i^{(\frac{5}{3} - 1)} \]Simplifying gives: \[ T_f = 300 \cdot 2^{-\frac{2}{3}} \approx 238 \text{ K} \]To find the final pressure, use \( P V^{\gamma} = \text{constant} \): \[ P_f = P_i \left( \frac{V_i}{V_f} \right)^\gamma = 1.00 \times 10^{5} \times (0.5)^\frac{5}{3} \approx 3.71 \times 10^4 \text{ Pa} \].

Step 5: Summary of Results

- Initial temperature: 300 K- Isothermal expansion: Final temperature = 300 K, Final pressure = \(5.00 \times 10^4 \) Pa- Isobaric expansion: Final temperature = 600 K, Final pressure = \(1.00 \times 10^5 \) Pa- Adiabatic expansion: Final temperature \( \approx 238 \) K, Final pressure \( \approx 3.71 \times 10^4 \) Pa

Key Concepts

Isothermal Process

An isothermal process is characterized by a constant temperature throughout the expansion or compression of a gas. Since the temperature remains unchanged, an isothermal process assumes that the internal energy of the gas is constant. This implies that any change in energy from work done by or on the gas is balanced by the heat exchange with the surroundings.

In our example with the ideal gas, during isothermal expansion, we kept the temperature at 300 K. Using the ideal gas law equation in the isothermal condition (\( PV = ext{constant} \)), it was determined that when the volume doubles, the pressure must halve. By applying this principle, we find the final pressure changes from \( 1.00 \times 10^{5} \) Pa to \( 5.00 \times 10^{4} \) Pa as the volume expands to twice its initial size.

Isobaric Process

In contrast to an isothermal process, an isobaric process maintains constant pressure. This constant pressure condition results in a direct change in volume and temperature. For an ideal gas under these conditions, the volume's increase necessitates a temperature increase to maintain the pressure constant.

With the isobaric process in our exercise, as the gas volume doubled, the temperature calculated using the ideal gas law also doubled. Thus, using \( PV = nRT \), the final temperature increased from 300 K to 600 K. Meanwhile, the pressure stays constant at \( 1.00 \times 10^{5} \) Pa. Understanding isobaric processes helps explain everyday phenomena, like how gases behave in piston engines.

Adiabatic Process

An adiabatic process uniquely signifies that no heat is exchanged with the environment. Here, changes in the gas's internal energy manifest solely due to work done on or by the system.

For a monatomic ideal gas, the adiabatic condition employs the relationship \( TV^{\gamma-1} = ext{constant} \) where \( \gamma = \frac{5}{3} \). During adiabatic expansion in our problem, the gas underwent cooling due to work done on the environment, resulting in the final temperature dropping to approximately 238 K. The pressure also decreases substantially from \( 1.00 \times 10^{5} \) Pa to around \( 3.71 \times 10^4 \) Pa. Such processes are critical in understanding natural phenomena, like the behavior of air parcels in meteorology.

Temperature Calculation

Calculating the temperature of a gas using the Ideal Gas Law is straightforward when the variables of pressure, volume, and the number of moles are known. The formula \( T = \frac{PV}{nR} \) allows for determining the temperature, which is essential in solving problems related to different gas processes.

For our initial condition, substituting the known values into the equation, i.e., \( P = 1.00 \times 10^{5} \) Pa, \( V = 2.50 \times 10^{-3} \) m³, \( n = 0.100 \) mol, and \( R = 8.314 \) J/mol·K, the initial temperature was found to be around 300 K. This calculation serves as a foundation for determining temperatures under varying conditions such as isothermal, isobaric, and adiabatic processes, helping us predict how a gas will behave when external conditions change.