Step-by-step Derivations

1. Identify the thermodynamic states:

• Initial State (State 1):

  • Pressure $P_i = 10^5 \, \text{Pa}$
  • Volume $V_i = 10^{-3} \, \text{m}^3$

• Final State (State 2):

  • Pressure $P_f = \left( \frac{1}{32} \right) \times 10^5 \, \text{Pa}$
  • Volume $V_f = 8 \times 10^{-3} \, \text{m}^3$

2. Analyze the two-step process: The system goes from the initial state to the final state via a two-step process:

• Step 1: Isobaric expansion (State 1 → State 3) The gas expands at a constant pressure $P_i$ from volume $V_i$ to $V_f$. Let the intermediate state be State 3.

  • $P_3 = P_i = 10^5 \, \text{Pa}$
  • $V_3 = V_f = 8 \times 10^{-3} \, \text{m}^3$

• Step 2: Isochoric process (State 3 → State 2) The gas is cooled at a constant volume $V_f$ until its pressure drops to $P_f$.

  • The process starts at State 3 $(P_3, V_3)$ and ends at the final State 2 $(P_f, V_f)$. This correctly connects the initial and final states.

3. Determine the properties of the gas: The problem states that an adiabatic process between the same initial and final states follows the relation $P^3V^5 = \text{constant}$. We assume this notation implies $P^3V^5$.

• Taking the cube root of both sides, we get $P V^{5/3} = \text{constant}$.
• For an adiabatic process, the governing equation is $P V^\gamma = \text{constant}$, where $\gamma$ is the adiabatic exponent (ratio of specific heats, $C_p/C_v$).
• Comparing the two equations, we find $\gamma = 5/3$.
• We know that $\gamma = 1 + \frac{2}{f}$, where f is the degrees of freedom. For $\gamma = 5/3$, f=3, which corresponds to a monatomic ideal gas.
• For a monatomic ideal gas, the molar specific heat at constant volume is $C_v = \frac{3}{2}R$, and at constant pressure is $C_p = \frac{5}{2}R$.

4. Calculate the total heat supplied (Q) for the two-step process: According to the First Law of Thermodynamics, the total heat supplied Q is the sum of the change in internal energy $\Delta U$ and the work done by the system W. $Q = \Delta U + W$ We will calculate $\Delta U$ and W for the overall process from State 1 to State 2 along the specified two-step path.

5. Calculate the total work done (W): The total work done is the sum of the work done in each step: $W = W_{1 \to 3} + W_{3 \to 2}$.

• Work in isobaric expansion ($W_{1 \to 3}$): $W_{1 \to 3} = P_i (V_f - V_i) = 10^5 \, \text{Pa} \times (8 \times 10^{-3} - 1 \times 10^{-3}) \, \text{m}^3$ $W_{1 \to 3} = 10^5 \times (7 \times 10^{-3}) = 700 \, \text{J}$
• Work in isochoric process ($W_{3 \to 2}$): Since the volume is constant (dV=0), the work done is zero. $W_{3 \to 2} = 0$
• Total work done: $W = 700 \, \text{J} + 0 = 700 \, \text{J}$

6. Calculate the change in internal energy (ΔU): Internal energy U is a state function, so its change $\Delta U$ depends only on the initial and final states, not the path taken.

• The change in internal energy for an ideal gas is given by $\Delta U = nC_v \Delta T$.
• Using the ideal gas law PV = nRT, we can write $\Delta U$ as: $\Delta U = U_f - U_i = nC_v(T_f - T_i) = \frac{C_v}{R}(nRT_f - nRT_i) = \frac{C_v}{R}(P_fV_f - P_iV_i)$
• We found $C_v = \frac{3}{2}R$, so $\frac{C_v}{R} = \frac{3}{2}$.
• Let's calculate the PV products:
  • $P_i V_i = (10^5 \, \text{Pa}) \times (10^{-3} \, \text{m}^3) = 100 \, \text{J}$
  • $P_f V_f = \left( \frac{1}{32} \times 10^5 \, \text{Pa} \right) \times (8 \times 10^{-3} \, \text{m}^3) = \frac{8}{32} \times 10^2 \, \text{J} = \frac{1}{4} \times 100 \, \text{J} = 25 \, \text{J}$
• Now, substitute these values into the $\Delta U$ equation: $\Delta U = \frac{3}{2} (25 \, \text{J} - 100 \, \text{J}) = \frac{3}{2} (-75 \, \text{J}) = -112.5 \, \text{J}$

7. Calculate the total heat supplied: Now we use the First Law of Thermodynamics: $Q = \Delta U + W = -112.5 \, \text{J} + 700 \, \text{J} = 587.5 \, \text{J}$

8. Conclusion: The calculated amount of heat supplied is $587.5 \, \text{J}$. This is approximately $588 \, \text{J}$. Comparing with the options:

• A: 112 J
• B: 294 J
• C: 588 J
• D: 813 J

The closest option is C.