Question

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 \(\times\) 10\(^5 \)Pa and occupies a volume of 2.50 \(\times\) 10\(^{-3}\) 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.

Short answer

(a) Initial temperature: 300.5 K. (b) Final conditions: (i) Isothermal: 300.5 K, 0.50 × 10⁵ Pa; (ii) Isobaric: 601.0 K, 1.00 × 10⁵ Pa; (iii) Adiabatic: 238.7 K, 0.316 × 10⁵ Pa.

Step-by-step solution

Understand Ideal Gas Law

The Ideal Gas Law is given by the equation \( PV = nRT \). Here, \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant (8.314 J/(mol·K)), and \( T \) is the temperature in Kelvin.

Solve for Initial Temperature

To find the initial temperature \( T_i \), rearrange the ideal gas law: \( T_i = \frac{PV}{nR} \). Plug in the given values: \( P = 1.00 \times 10^5 \) Pa, \( V = 2.50 \times 10^{-3} \) m\(^3\), \( n = 0.100 \) mol, and \( R = 8.314 \) J/(mol·K), to get \( T_i = \frac{(1.00 \times 10^5)(2.50 \times 10^{-3})}{0.100 \, \cdot 8.314} \).

Calculate Initial Temperature

Perform the calculations: \( T_i = \frac{2.5 \times 10^2}{0.8314} \approx 300.5 \) K.

Solve Isothermal Expansion (i)

For isothermal expansion, the temperature remains constant. Therefore, \( T_f = T_i = 300.5 \) K. Using \( P_i V_i = P_f V_f \), rearrange to find \( P_f \): \( P_f = \frac{P_i V_i}{V_f} \). Substitute \( V_f = 2 \times 2.50 \times 10^{-3} = 5.00 \times 10^{-3} \) m\(^3\) to get \( P_f = \frac{1.00 \times 10^5 \cdot 2.50 \times 10^{-3}}{5.00 \times 10^{-3}} = 0.50 \times 10^5 \) Pa.

Solve Isobaric Expansion (ii)

For an isobaric process, the pressure remains constant. So \( P_f = P_i = 1.00 \times 10^5 \) Pa. Using the ideal gas law \( \frac{T_f}{T_i} = \frac{V_f}{V_i} \), calculate \( T_f = T_i \cdot \frac{V_f}{V_i} = 300.5 \cdot \frac{5.00 \times 10^{-3}}{2.50 \times 10^{-3}} = 601.0 \) K.

Solve Adiabatic Expansion (iii)

For a monatomic gas, the adiabatic condition is given by \( PV^\gamma = \text{constant} \), where \( \gamma = \frac{5}{3} \). Using \( T_f = T_i \cdot \left(\frac{V_i}{V_f}\right)^{\gamma-1} \), calculate \( T_f = 300.5 \cdot \left(\frac{2.50 \times 10^{-3}}{5.00 \times 10^{-3}}\right)^{2/3} \approx 238.7 \) K. For \( P_f \), use \( P_f = P_i \cdot \left(\frac{V_i}{V_f}\right)^\gamma \approx 0.316 \times 10^5 \) Pa.

Key concepts

Isothermal Process

In an isothermal process, the temperature remains constant throughout the expansion or compression of the gas. This means, that for an isothermal expansion, the initial and final temperatures are equal. According to the ideal gas law, when temperature is constant,

• The product of pressure (\(P\)) and volume (\(V\)) remains constant: \(PV = ext{constant}\).
• For the expansion of the gas to twice its initial volume, \(V_f = 2V_i\).
• Using this property, we can find the final pressure:\(P_f = \frac{P_i V_i}{V_f}\).

The most important aspect of the isothermal process is the constancy of temperature; hence, internal energy changes only through work done by or on the gas. This process requires perfect thermal equilibrium with the surroundings to keep the temperature unchanged.

Isobaric Process

In an isobaric process, the pressure remains constant. This phenomenon affects other gas properties as they change according to the ideal gas law. Since the pressure does not change:

• The relationship between temperature and volume becomes directly proportional.
• Using the equation \(\frac{T_f}{T_i} = \frac{V_f}{V_i}\), it shows that the final temperature is determined by the change in volume.

For the gas to expand to twice its initial volume, the ideal gas law simplifies the calculation for the final state properties given constant pressure. During such processes, thermal energy is absorbed or released to maintain constant pressure, leading to a change in the gas's volume and temperature.

Adiabatic Expansion

Adiabatic processes are distinctive because they occur without the transfer of heat between the system and its surroundings. For a monatomic ideal gas, the key equations include:

• The adiabatic equation: \(PV^\gamma = ext{constant}\), where \(\gamma = \frac{5}{3}\) (ratio of specific heats for a monatomic gas).
• Temperature changes even though no heat enters or leaves the system.
• Final temperature can be determined with \(T_f = T_i \left(\frac{V_i}{V_f}\right)^{\gamma-1}\).

Since adiabatic processes occur without heat exchange, changes in internal energy are driven solely by work done on or by the gas. This process is significant in understanding how temperature and pressure can drastically change together in closed systems.