Question
Part 1: An ideal gas at temperature T0 is slowly compressed at constant pressure of 2 atm from a volume of 10 liters to a volume of 2 liters. Then the volume of the gas is held constant while heat is added, raising the gas temperature back to T0. Calculate the work done ON the gas. 1 atm = 1.0×10^5 Pascals and 1 liter = 0.001 m^3.
1. +800 J
2. −400 J
3. +400 J
4. −800 J
5. −1600 J
6. +1600 J
Part 2: Calculate the heat flow INTO the gas
1. −800 J
2. −1600 J
3. −400 J
4. +800 J
5. +1600 J
6. +400 J
1. +800 J
2. −400 J
3. +400 J
4. −800 J
5. −1600 J
6. +1600 J
Part 2: Calculate the heat flow INTO the gas
1. −800 J
2. −1600 J
3. −400 J
4. +800 J
5. +1600 J
6. +400 J
Answers
GPT-4o mini
To solve this problem, let's analyze each part step by step.
### Part 1: Work Done ON the Gas
The work done on the gas during a compression process at constant pressure can be calculated using the formula:
\[
W = P \Delta V
\]
where:
- \(W\) = work done (in Joules)
- \(P\) = pressure (in Pascals)
- \(\Delta V\) = change in volume (in cubic meters)
1. **Compute the change in volume**:
- Initial volume, \(V_i = 10 \, \text{liters} = 10 \times 0.001 \, \text{m}^3 = 0.01 \, \text{m}^3\)
- Final volume, \(V_f = 2 \, \text{liters} = 2 \times 0.001 \, \text{m}^3 = 0.002 \, \text{m}^3\)
- Change in volume, \(\Delta V = V_f - V_i = 0.002 \, \text{m}^3 - 0.01 \, \text{m}^3 = -0.008 \, \text{m}^3\)
2. **Convert the pressure to Pascals**:
- The constant pressure is given as \(2 \, \text{atm} = 2 \times 1.0 \times 10^5 \, \text{Pa} = 2.0 \times 10^5 \, \text{Pa}\)
3. **Calculate the work done **:
\[
W = (2.0 \times 10^5 \, \text{Pa}) \times (-0.008 \, \text{m}^3) = -1600 \, \text{J}
\]
So, the work done ON the gas is **−1600 J** (the negative sign indicates work is done on the gas).
**Part 1 Answer: 5. −1600 J**
### Part 2: Heat Flow INTO the Gas
To determine the heat flow into the gas, we can apply the First Law of Thermodynamics:
\[
\Delta U = Q - W
\]
Where:
- \(\Delta U\) = change in internal energy
- \(Q\) = heat added to the system
- \(W\) = work done on the system (this is already calculated)
When the gas is compressed and then heated back to the original temperature \(T_0\), we can consider that if the gas returns to its original temperature, the change in internal energy \(\Delta U\) is zero (since internal energy for an ideal gas only depends on temperature).
Thus:
\[
0 = Q - (-1600 \, J)
\]
Therefore:
\[
Q = 1600 \, J
\]
Since this is heat added to the gas:
**Part 2 Answer: 5. +1600 J**
### Part 1: Work Done ON the Gas
The work done on the gas during a compression process at constant pressure can be calculated using the formula:
\[
W = P \Delta V
\]
where:
- \(W\) = work done (in Joules)
- \(P\) = pressure (in Pascals)
- \(\Delta V\) = change in volume (in cubic meters)
1. **Compute the change in volume**:
- Initial volume, \(V_i = 10 \, \text{liters} = 10 \times 0.001 \, \text{m}^3 = 0.01 \, \text{m}^3\)
- Final volume, \(V_f = 2 \, \text{liters} = 2 \times 0.001 \, \text{m}^3 = 0.002 \, \text{m}^3\)
- Change in volume, \(\Delta V = V_f - V_i = 0.002 \, \text{m}^3 - 0.01 \, \text{m}^3 = -0.008 \, \text{m}^3\)
2. **Convert the pressure to Pascals**:
- The constant pressure is given as \(2 \, \text{atm} = 2 \times 1.0 \times 10^5 \, \text{Pa} = 2.0 \times 10^5 \, \text{Pa}\)
3. **Calculate the work done **:
\[
W = (2.0 \times 10^5 \, \text{Pa}) \times (-0.008 \, \text{m}^3) = -1600 \, \text{J}
\]
So, the work done ON the gas is **−1600 J** (the negative sign indicates work is done on the gas).
**Part 1 Answer: 5. −1600 J**
### Part 2: Heat Flow INTO the Gas
To determine the heat flow into the gas, we can apply the First Law of Thermodynamics:
\[
\Delta U = Q - W
\]
Where:
- \(\Delta U\) = change in internal energy
- \(Q\) = heat added to the system
- \(W\) = work done on the system (this is already calculated)
When the gas is compressed and then heated back to the original temperature \(T_0\), we can consider that if the gas returns to its original temperature, the change in internal energy \(\Delta U\) is zero (since internal energy for an ideal gas only depends on temperature).
Thus:
\[
0 = Q - (-1600 \, J)
\]
Therefore:
\[
Q = 1600 \, J
\]
Since this is heat added to the gas:
**Part 2 Answer: 5. +1600 J**