A system is undergoing a cycle that consists of three processes. During the first process, the work is 5 kJ and the heat is 23 kJ. In the second process no work takes place and the heat interaction is -50 kJ. The third process is adiabatic. The work in the third process is
-32 kJ
Thermodynamics Cycle Analysis
This question involves analyzing a thermodynamic cycle composed of three distinct processes. We need to determine the work done during the third process, given information about heat and work interactions in the first two processes and the nature of the third process.
Understanding the First Law of Thermodynamics
The core principle governing this problem is the First Law of Thermodynamics. For a complete thermodynamic cycle, the total change in internal energy ($\Delta U$) is zero. The First Law states that the net heat transfer ($Q_{net}$) to the system equals the net work done ($W_{net}$) by the system.
Mathematically, for a cycle:
$$ \Delta U_{cycle} = 0 $$And using the First Law for each process ($Q = \Delta U + W$):
$$ Q_{net} = \sum Q = \sum (\Delta U + W) = \sum \Delta U + \sum W $$Since $\sum \Delta U = \Delta U_{cycle} = 0$, we have:
$$ Q_{net} = W_{net} $$This fundamental relationship will be used to solve the problem.
Step-by-Step Calculation
Let's break down the cycle process by process:
- Process 1:
- Work ($W_1$) = 5 kJ
- Heat ($Q_1$) = 23 kJ
- Process 2:
- Work ($W_2$) = 0 kJ (no work takes place)
- Heat ($Q_2$) = -50 kJ
- Process 3:
- Adiabatic process, meaning Heat ($Q_3$) = 0 kJ
- Work ($W_3$) = ? (This is what we need to find)
Calculating Net Heat and Net Work
First, we calculate the total heat transfer across all three processes:
$$ Q_{net} = Q_1 + Q_2 + Q_3 $$ $$ Q_{net} = 23 \text{ kJ} + (-50 \text{ kJ}) + 0 \text{ kJ} $$ $$ Q_{net} = 23 \text{ kJ} - 50 \text{ kJ} $$ $$ Q_{net} = -27 \text{ kJ} $$Next, we express the total work done across all three processes:
$$ W_{net} = W_1 + W_2 + W_3 $$ $$ W_{net} = 5 \text{ kJ} + 0 \text{ kJ} + W_3 $$ $$ W_{net} = 5 \text{ kJ} + W_3 $$Determining Work in the Third Process
Now, we apply the cycle condition ($Q_{net} = W_{net}$):
$$ -27 \text{ kJ} = 5 \text{ kJ} + W_3 $$To find $W_3$, we rearrange the equation:
$$ W_3 = -27 \text{ kJ} - 5 \text{ kJ} $$ $$ W_3 = -32 \text{ kJ} $$Therefore, the work done in the third process is -32 kJ.
Final Answer Verification
The calculated work for the third process ($W_3$) is -32 kJ. This matches one of the provided options.
Important Questions from Statement of First Law of Thermodynamics
Which term of the following is defined by the first law of thermodynamics?
- A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 0.105 MPa to a final state of 0.15 m3 and 0.105 MPa, the pressure remaining constant during the process. There is a transfer of 37.6 kJ of heat from the gas during the process. How much does the internal energy of the gas change?
The first law of thermodynamics is synonymous with the ______.
Which one of the following process is the energy efficient one for compression of air
The first law of thermodynamics mentions the ______.
