Question
An engine using 1 mol of an ideal gas initially at 15.4 L and 314 K performs a cycle consisting of four steps:
1) an isothermal expansion at 314 K from 15.4 L to 35 L
2) cooling at constant volume to 162 K
3) an isothermal compression to its original volume of 15.4 Land
4) heating at constant volume to its original temperature of 314 K. Find its efficiency. Assume that the heat capacity is 21 J/K and the universal gas constant is 0.08206 L · atm/mol/K =8.314 J/mol/K. Give your answer in the form of a percentage.
1) an isothermal expansion at 314 K from 15.4 L to 35 L
2) cooling at constant volume to 162 K
3) an isothermal compression to its original volume of 15.4 Land
4) heating at constant volume to its original temperature of 314 K. Find its efficiency. Assume that the heat capacity is 21 J/K and the universal gas constant is 0.08206 L · atm/mol/K =8.314 J/mol/K. Give your answer in the form of a percentage.
Answers
GPT-4o mini
To calculate the efficiency of the engine using 1 mol of an ideal gas through the described cycle, we can determine the work done during the cycles and the heat absorbed/released during the processes. We first summarize the steps and their specific calculations.
### Step 1: Isothermal Expansion at 314 K
For isothermal processes, the work done \( W \) can be calculated as:
\[
W = nRT \ln \left( \frac{V_f}{V_i} \right)
\]
Given:
- \( n = 1 \, \text{mol} \)
- \( R = 8.314 \, \text{J/(mol·K)} \)
- \( T = 314 \, \text{K} \)
- \( V_i = 15.4 \, \text{L} \)
- \( V_f = 35 \, \text{L} \)
Calculating the work done during isothermal expansion:
\[
W_{12} = 1 \times 8.314 \times 314 \times \ln \left( \frac{35}{15.4} \right)
\]
Calculating \( \ln \left( \frac{35}{15.4} \right) \):
\[
\frac{35}{15.4} \approx 2.273
\]
\[
\ln(2.273) \approx 0.827
\]
Thus:
\[
W_{12} \approx 1 \times 8.314 \times 314 \times 0.827 \approx 2432.44 \, \text{J}
\]
### Step 2: Cooling at Constant Volume to 162 K
At constant volume, the change in internal energy \( \Delta U \) is directly related to the heat loss:
\[
Q_{21} = nC_v \Delta T
\]
Where \( C_v \) can be derived from \( C_p \) for ideal monatomic gases, which is:
\[
C_v = C_p - R
\]
Let's assume \( C_p = 21 \, \text{J/K} \) (as given). Therefore:
\[
C_v = 21 - 8.314 \approx 12.686 \, \text{J/K}
\]
Calculating heat removed while cooling:
\[
Q_{21} = 1 \times 12.686 \, \text{J/K} \times (314 - 162) \, \text{K}
\]
\[
= 12.686 \times 152 \approx 1925.52 \, \text{J}
\]
### Step 3: Isothermal Compression Back to 15.4 L
For isothermal compression, the work done on the gas is of the same magnitude as the work done on the gas in Step 1 but opposite in sign:
\[
W_{34} = -W_{12} \approx -2432.44 \, \text{J}
\]
### Step 4: Heating at Constant Volume to 314 K
Heat added during heating at constant volume is:
\[
Q_{43} = nC_v \Delta T
\]
Calculating heat added while heating:
\[
Q_{43} = 1 \times 12.686 \, \text{J/K} \times (314 - 162) \, \text{K}
\]
\[
= 12.686 \times 152 \approx 1925.52 \, \text{J}
\]
### Total Heat Exchange
Total heat absorbed during the cycle:
\[
Q_{\text{in}} = Q_{12} + Q_{43} = 2432.44 + 1925.52 = 4357.96 \, \text{J}
\]
Total heat rejected during the cycle:
\[
Q_{\text{out}} = Q_{21} + W_{34} = 1925.52 - 2432.44 = -506.92 \, \text{J}
\]
### Efficiency of the Cycle
Finally, we can find the efficiency \( \eta \):
\[
\eta = \frac{Q_{\text{in}} - |Q_{\text{out}}|}{Q_{\text{in}}}
\]
Where \(|Q_{\text{out}}| = 506.92 \, \text{J} \).
Calculating:
\[
\eta = \frac{4357.96 - 506.92}{4357.96}
\]
\[
\eta = \frac{3851.04}{4357.96} \approx 0.884
\]
Converting to percentage:
\[
\text{Efficiency} \approx 88.4\%
\]
Thus, the efficiency of the engine is approximately **88.4%**.
### Step 1: Isothermal Expansion at 314 K
For isothermal processes, the work done \( W \) can be calculated as:
\[
W = nRT \ln \left( \frac{V_f}{V_i} \right)
\]
Given:
- \( n = 1 \, \text{mol} \)
- \( R = 8.314 \, \text{J/(mol·K)} \)
- \( T = 314 \, \text{K} \)
- \( V_i = 15.4 \, \text{L} \)
- \( V_f = 35 \, \text{L} \)
Calculating the work done during isothermal expansion:
\[
W_{12} = 1 \times 8.314 \times 314 \times \ln \left( \frac{35}{15.4} \right)
\]
Calculating \( \ln \left( \frac{35}{15.4} \right) \):
\[
\frac{35}{15.4} \approx 2.273
\]
\[
\ln(2.273) \approx 0.827
\]
Thus:
\[
W_{12} \approx 1 \times 8.314 \times 314 \times 0.827 \approx 2432.44 \, \text{J}
\]
### Step 2: Cooling at Constant Volume to 162 K
At constant volume, the change in internal energy \( \Delta U \) is directly related to the heat loss:
\[
Q_{21} = nC_v \Delta T
\]
Where \( C_v \) can be derived from \( C_p \) for ideal monatomic gases, which is:
\[
C_v = C_p - R
\]
Let's assume \( C_p = 21 \, \text{J/K} \) (as given). Therefore:
\[
C_v = 21 - 8.314 \approx 12.686 \, \text{J/K}
\]
Calculating heat removed while cooling:
\[
Q_{21} = 1 \times 12.686 \, \text{J/K} \times (314 - 162) \, \text{K}
\]
\[
= 12.686 \times 152 \approx 1925.52 \, \text{J}
\]
### Step 3: Isothermal Compression Back to 15.4 L
For isothermal compression, the work done on the gas is of the same magnitude as the work done on the gas in Step 1 but opposite in sign:
\[
W_{34} = -W_{12} \approx -2432.44 \, \text{J}
\]
### Step 4: Heating at Constant Volume to 314 K
Heat added during heating at constant volume is:
\[
Q_{43} = nC_v \Delta T
\]
Calculating heat added while heating:
\[
Q_{43} = 1 \times 12.686 \, \text{J/K} \times (314 - 162) \, \text{K}
\]
\[
= 12.686 \times 152 \approx 1925.52 \, \text{J}
\]
### Total Heat Exchange
Total heat absorbed during the cycle:
\[
Q_{\text{in}} = Q_{12} + Q_{43} = 2432.44 + 1925.52 = 4357.96 \, \text{J}
\]
Total heat rejected during the cycle:
\[
Q_{\text{out}} = Q_{21} + W_{34} = 1925.52 - 2432.44 = -506.92 \, \text{J}
\]
### Efficiency of the Cycle
Finally, we can find the efficiency \( \eta \):
\[
\eta = \frac{Q_{\text{in}} - |Q_{\text{out}}|}{Q_{\text{in}}}
\]
Where \(|Q_{\text{out}}| = 506.92 \, \text{J} \).
Calculating:
\[
\eta = \frac{4357.96 - 506.92}{4357.96}
\]
\[
\eta = \frac{3851.04}{4357.96} \approx 0.884
\]
Converting to percentage:
\[
\text{Efficiency} \approx 88.4\%
\]
Thus, the efficiency of the engine is approximately **88.4%**.