Question
During an adiabatic expansion the temperature of 0.600mol of oxygen drops from 30°C to 10°C.
1.How much work does the gas do?
2.How much heat is added to the gas?
1.How much work does the gas do?
2.How much heat is added to the gas?
Answers
GPT-4o mini
To analyze the adiabatic expansion of the gas, we need to apply the relevant thermodynamic principles.
1. **Work Done by the Gas (W):**
For an adiabatic process, the work done by the gas can be calculated using the first law of thermodynamics, which states:
\[
\Delta U = Q - W
\]
Where:
- \(\Delta U\) is the change in internal energy.
- \(Q\) is the heat added to the system (which is 0 for an adiabatic process).
- \(W\) is the work done by the system.
Since the process is adiabatic, \(Q = 0\), so the equation simplifies to:
\[
\Delta U = -W
\]
To find the change in internal energy, we use the formula:
\[
\Delta U = nC_v \Delta T
\]
Where:
- \(n\) is the number of moles (0.600 mol).
- \(C_v\) is the molar heat capacity at constant volume (for diatomic gases like O2, \(C_v \approx 5/2 R\) where \(R = 8.314 \, \text{J/(mol·K)}\)).
- \(\Delta T\) is the change in temperature.
First, let's convert the temperatures from Celsius to Kelvin:
- \(T_i = 30°C = 303.15 \, K\)
- \(T_f = 10°C = 283.15 \, K\)
Now, calculate \(\Delta T\):
\[
\Delta T = T_f - T_i = 283.15 \, K - 303.15 \, K = -20 \, K
\]
Now, we calculate \(\Delta U\):
\[
C_v = \frac{5}{2} R = \frac{5}{2} \times 8.314 \, \text{J/(mol·K)} \approx 20.786 \, \text{J/(mol·K)}
\]
\[
\Delta U = nC_v \Delta T = 0.600 \, \text{mol} \times 20.786 \, \text{J/(mol·K)} \times (-20 \, K)
\]
Calculating \(\Delta U\):
\[
\Delta U = 0.600 \times 20.786 \times (-20) \approx -249.432\, \text{J}
\]
Since \(\Delta U = -W\):
\[
W = -\Delta U = 249.432\, \text{J}
\]
Thus, the work done by the gas is approximately:
\[
W \approx 249.43 \, \text{J}
\]
2. **Heat Added to the Gas (Q):**
As already established, for an adiabatic process, no heat is added or removed from the system:
\[
Q = 0 \, \text{J}
\]
### Summary:
1. The work done by the gas is approximately **249.43 J**.
2. The heat added to the gas is **0 J** (since the process is adiabatic).
1. **Work Done by the Gas (W):**
For an adiabatic process, the work done by the gas can be calculated using the first law of thermodynamics, which states:
\[
\Delta U = Q - W
\]
Where:
- \(\Delta U\) is the change in internal energy.
- \(Q\) is the heat added to the system (which is 0 for an adiabatic process).
- \(W\) is the work done by the system.
Since the process is adiabatic, \(Q = 0\), so the equation simplifies to:
\[
\Delta U = -W
\]
To find the change in internal energy, we use the formula:
\[
\Delta U = nC_v \Delta T
\]
Where:
- \(n\) is the number of moles (0.600 mol).
- \(C_v\) is the molar heat capacity at constant volume (for diatomic gases like O2, \(C_v \approx 5/2 R\) where \(R = 8.314 \, \text{J/(mol·K)}\)).
- \(\Delta T\) is the change in temperature.
First, let's convert the temperatures from Celsius to Kelvin:
- \(T_i = 30°C = 303.15 \, K\)
- \(T_f = 10°C = 283.15 \, K\)
Now, calculate \(\Delta T\):
\[
\Delta T = T_f - T_i = 283.15 \, K - 303.15 \, K = -20 \, K
\]
Now, we calculate \(\Delta U\):
\[
C_v = \frac{5}{2} R = \frac{5}{2} \times 8.314 \, \text{J/(mol·K)} \approx 20.786 \, \text{J/(mol·K)}
\]
\[
\Delta U = nC_v \Delta T = 0.600 \, \text{mol} \times 20.786 \, \text{J/(mol·K)} \times (-20 \, K)
\]
Calculating \(\Delta U\):
\[
\Delta U = 0.600 \times 20.786 \times (-20) \approx -249.432\, \text{J}
\]
Since \(\Delta U = -W\):
\[
W = -\Delta U = 249.432\, \text{J}
\]
Thus, the work done by the gas is approximately:
\[
W \approx 249.43 \, \text{J}
\]
2. **Heat Added to the Gas (Q):**
As already established, for an adiabatic process, no heat is added or removed from the system:
\[
Q = 0 \, \text{J}
\]
### Summary:
1. The work done by the gas is approximately **249.43 J**.
2. The heat added to the gas is **0 J** (since the process is adiabatic).