Problem 52 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Two moles of a perfect gas undergo the following processes: (a) a reversible isobaric expansion from \((1.0 \mathrm{~atm}, 20.0 \mathrm{~L})\) to \((1.0 \mathrm{~atm}, 40.0\) L); (b) a reversible isochoric change of state from \((1.0 \mathrm{~atm}, 40.0 \mathrm{~L})\) to \((0.5\) atm, \(40.0 \mathrm{~L}\) ); (c) a reversible isothermal compression from \((0.5 \mathrm{~atm}, 40.0 \mathrm{~L})\) to \((1.0\) atm, \(20.0 \mathrm{~L}\) ). (i) Sketch with labels each of the processes on the same \(P-V\) diagram. (ii) Calculate the total work \((W)\) and the total heat change \((q)\) involved in the above processes. (iii) What will be the values of \(\Delta U, \Delta H\) and \(\Delta S\) for the overall process?

Step-by-Step Solution

Answer

Sketch processes on a PV diagram, calculate work, heat, \\Delta U, \\Delta H, and \\Delta S for each and in total using formulas for each process type.

Step 1: Understand the Processes

Consider three processes: (a) isobaric expansion, (b) isochoric change, and (c) isothermal compression. Note that each process affects the properties like pressure (P), volume (V), and temperature (T) differently for a perfect gas.

Step 2: Sketch the PV Diagram

For (i), sketch the processes: - Process (a) is a horizontal line from 20.0 L to 40.0 L at 1.0 atm. - Process (b) is a vertical line down from 1.0 atm to 0.5 atm at 40.0 L. - Process (c) is a curve from 0.5 atm, 40.0 L to 1.0 atm, 20.0 L. Ensure each process is labeled correctly on the diagram.

Step 3: Calculate Work Done (Process (a))

For process (a) (isobaric expansion), calculate work done using \[ W_a = P \Delta V = 1.0 \text{ atm} \times (40.0 - 20.0) \text{ L} = 20.0 \text{ L atm} \].Convert to Joules: 1 atm L = 101.325 J, so \[ W_a = 20.0 \times 101.325 = 2026.5 \text{ J} \].

Step 4: Calculate Heat Change (Process (a))

Use the formula for heat in isobaric process: \[ q_a = nC_p \Delta T \].With 2 moles, assume ideal gas and calculate:- Use the ideal gas law to find \( \Delta T \) based on \( PV = nRT \).- \( C_p = \frac{5}{2} R \); calculate \( q_a = 2 \times \frac{5}{2} R \Delta T \).

Step 5: Calculate Work Done (Process (b))

For process (b) (isochoric), no work is done since volume does not change: \[ W_b = 0 \].

Step 6: Calculate Heat Change (Process (b))

For isochoric processes, use \[ q_b = nC_v \Delta T \].Since no work is done, find \( q_b \) through calculations similar to Step 4, using \( C_v = \frac{3}{2} R \).

Step 7: Calculate Work Done (Process (c))

For process (c) (isothermal compression), calculate work done using \[ W_c = nRT \ln \left(\frac{V_i}{V_f}\right) \]Where \( V_i = 40.0 \text{ L}, V_f = 20.0 \text{ L} \). Use \( R \) and previously determined \( T \).

Step 8: Calculate Heat Change (Process (c))

Since the process is isothermal, the change in internal energy \( \Delta U = 0 \). Therefore, \( q_c = W_c \) for process (c).

Step 9: Calculate Total Work and Total Heat

Sum up the work and heat from each process:\[ W_{total} = W_a + W_b + W_c \]\[ q_{total} = q_a + q_b + q_c \].

Step 10: Calculate Change in Internal Energy (\(\Delta U\))

The total change in internal energy \( \Delta U \) is given by the first law of thermodynamics:\[ \Delta U = q_{total} - W_{total} \].

Step 11: Calculate Enthalpy Change (\(\Delta H\))

For the overall process involving constant pressure and volume changes, use:\[ \Delta H = \Delta U + \Delta (PV) \].Consider \( \Delta (PV) \) by evaluating PV at initial and final states.

Step 12: Calculate Entropy Change (\(\Delta S\))

Calculate entropy changes for each process and sum them:Isobaric: \( \Delta S_a = nC_p \ln \left(\frac{T_f}{T_i}\right) \),Isochoric: \( \Delta S_b = nC_v \ln \left(\frac{T_f}{T_i}\right) \),Isothermal: \( \Delta S_c = \frac{q_c}{T} \).

Step 13: Summarize the Overall Changes

Summarize the calculations:- Total work \( W_{total} \)- Total heat \( q_{total} \)- \( \Delta U \), \( \Delta H \), and \( \Delta S \).

Key Concepts

Isobaric process

An isobaric process refers to a thermodynamic transformation that occurs at constant pressure. This means that no matter how the volume changes, the pressure remains constant throughout the process. In this kind of process, energy in the form of heat is either added to or removed from the system. It's important to correctly calculate this energy change as it affects the work done by the system.

During an isobaric expansion, such as moving from 20.0 L to 40.0 L, the gas does work on the surroundings. The work done is calculated using the formula: \[ W = P \Delta V \] where \( P \) is the pressure and \( \Delta V \) is the change in volume. This process illustrates how gases can perform work by expanding against external pressures while maintaining a stable pressure internally.

In isobaric processes, the heat involved \( q \) is based on the specific heat capacity at constant pressure \( C_p \) and the change in temperature \( \Delta T \). Using the equation \[ q = nC_p \Delta T \] where \( n \) is the number of moles, we can find the heat absorbed or released. This directly impacts the energy state of the system.

Isochoric process

An isochoric process is characterized by constant volume, meaning that no work is done by or on the system, as there is no volume change. In simple terms, even if the pressure and temperature change during the process, the volume stays the same.

For example, when the state changes from \( 1.0 \text{ atm}, 40.0 \text{ L} \) to \( 0.5 \text{ atm}, 40.0 \text{ L} \), the volume remains constant at 40.0 L. This results in no work being done, as \( W = P \Delta V = 0 \).

However, heat transfer \( q \) in an isochoric process is still possible and necessary for the change of state of the system. It's calculated using the formula: \[ q = nC_v \Delta T \] where \( C_v \) is the specific heat at constant volume. This heat change directly affects the internal energy \( \Delta U \) because, according to the first law of thermodynamics, the change in internal energy equals the heat added: \[ \Delta U = q \].In isochoric processes, the absence of work makes it easier to analyze energy changes, focusing directly on heat exchanges.

Isothermal process

An isothermal process is one where the temperature of the system remains constant throughout the process. This means any heat added to the system is used to do work, resulting in no change to the internal energy \( \Delta U \) of the system: \[ \Delta U = 0 \].Isothermal processes require a perfect balance between heat exchange and work done to maintain the constant temperature.

In our problem, the isothermal compression takes the system from \( 0.5 \text{ atm}, 40.0 \text{ L} \) to \( 1.0 \text{ atm}, 20.0 \text{ L} \). The work done \( W \) during isothermal processes can be complex to determine because it involves calculating how the volume changes at a constant temperature: \[ W = nRT \ln \left(\frac{V_i}{V_f}\right) \] where \( V_i \) and \( V_f \) represent initial and final volumes, and \( R \) is the ideal gas constant.

Heat \( q \) in isothermal processes is equal in magnitude to the work done, since the temperature stays the same and there is no change in internal energy. Therefore, according to the first law of thermodynamics, this means: \[ q = W \].Isothermal processes beautifully illustrate the masterful equilibrium of thermodynamic forces, including heat, work, and energy conservation.