Question

In a polytropic process, heat rejected is given by

• A. \( \frac{\gamma-n}{\gamma-1} \times \) Work done on the system
• B. \( \frac{\gamma}{\gamma-1} \times \) Work done on the system
• C. \( \frac{\gamma-n}{\gamma} \times \) Work done on the system
• D. \( \frac{\gamma-n}{n} \times \) Work done on the system

Hint:

For a polytropic process (\(PV^n = \text{constant}\)) with an ideal gas, the heat transferred can be related to the work done. The heat rejected is given by the formula \( \text{Heat rejected} = \frac{\gamma - n}{\gamma - 1} \times \text{Work done on the system} \). This is a common derivation in thermodynamics.

Answer 1

The Correct Option is A

Step 1: Recall the First Law of Thermodynamics for a process.
The First Law of Thermodynamics states that the heat added to a system (\(Q\)) is equal to the change in internal energy (\(\Delta U\)) plus the work done by the system (\(W\)): \[ Q = \Delta U + W \] If \(W_{on}\) is the work done on the system, then \(W = -W_{on}\). So, \(Q = \Delta U - W_{on}\).
Step 2: Express change in internal energy and work done for a polytropic process.
For an ideal gas, the change in internal energy is given by: \[ \Delta U = m c_v (T_2 - T_1) \] We know that \( c_v = \frac{R}{\gamma - 1} \), so \( \Delta U = \frac{mR(T_2 - T_1)}{\gamma - 1} \). Since \( mR(T_2 - T_1) = P_2 V_2 - P_1 V_1 \) for an ideal gas, we have: \[ \Delta U = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} \] The work done by the system in a polytropic process (\(PV^n = C\)) is: \[ W = \frac{P_1 V_1 - P_2 V_2}{n - 1} \quad \text{for } n \neq 1 \] Therefore, the work done on the system is \(W_{on} = -W = \frac{P_2 V_2 - P_1 V_1}{n - 1}\).
From this, we can write \( P_2 V_2 - P_1 V_1 = (n - 1) W_{on} \).
Step 3: Derive the heat rejected (or transferred).
Substitute the expressions for \(\Delta U\) and \(W_{on}\) into the First Law \(Q = \Delta U - W_{on}\): \[ Q = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1} - W_{on} \] Now substitute \( P_2 V_2 - P_1 V_1 = (n - 1) W_{on} \): \[ Q = \frac{(n - 1) W_{on}}{\gamma - 1} - W_{on} \] Factor out \(W_{on}\): \[ Q = W_{on} \left( \frac{n - 1}{\gamma - 1} - 1 \right) \] \[ Q = W_{on} \left( \frac{n - 1 - (\gamma - 1)}{\gamma - 1} \right) \] \[ Q = W_{on} \left( \frac{n - \gamma}{\gamma - 1} \right) \] The question asks for "heat rejected". If \(Q\) is positive, it means heat added. If \(Q\) is negative, it means heat rejected. So, "heat rejected" is \( -Q \). \[ \text{Heat rejected} = -Q = -W_{on} \left( \frac{n - \gamma}{\gamma - 1} \right) \] \[ \text{Heat rejected} = W_{on} \left( \frac{\gamma - n}{\gamma - 1} \right) \] This matches option (1). The final answer is \( \boxed{\text{1}} \).