To solve the problem, we need to apply the relationships between pressure, volume, temperature, and polytropic processes. Let's go through the calculations step by step:
Given Data:
- Mass of the gas, \( m = 1 , \text{kg} \)
- Initial pressure, \( P_1 = 1300 , \text{kPa} = 1300 \times 10^3 , \text{Pa} \)
- Initial temperature, \( T_1 = 250 , \text{°C} = 250 + 273.15 = 523.15 , \text{K} \)
- Final volume, \( V_2 = 2 , \text{m}^3 \)
- Final temperature, \( T_2 = 37 , \text{°C} = 37 + 273.15 = 310.15 , \text{K} \)
- Gas constant, \( R = 0.287 , \text{kJ/kg K} = 287 , \text{J/kg K} \)
2.1 Final Pressure after Expansion
Using the ideal gas equation:
\[ P = \frac{mRT}{V} \]
We can find the final pressure \( P_2 \) at the final state:
\[ P_2 = \frac{mRT_2}{V_2} \]
Substituting in the values:
\[ P_2 = \frac{(1 , \text{kg}) (287 , \text{J/kg K}) (310.15 , \text{K})}{2 , \text{m}^3} \]
Calculating:
\[ P_2 = \frac{287 \times 310.15}{2} = \frac{89119.05}{2} = 44559.525 , \text{Pa} \approx 44.56 , \text{kPa} \]
2.2 Index of Expansion (\( n \))
For a polytropic process, we have the relation for pressures and specific volumes:
\[ P_1 V_1^n = P_2 V_2^n \implies \left( \frac{P_1}{P_2} \right) = \left( \frac{V_2}{V_1} \right)^n \]
First, we need to find \( V_1 \) in the next step. We can find the initial volume using the ideal gas law.
2.3 Initial Volume (\( V_1 \))
Rearranging the ideal gas law for the initial state:
\[ V_1 = \frac{mRT_1}{P_1} \]
Substituting in the values:
\[ V_1 = \frac{(1 , \text{kg}) (287 , \text{J/kg K}) (523.15 , \text{K})}{1300 \times 10^3 , \text{Pa}} \]
Calculating:
\[ V_1 = \frac{287 \times 523.15}{1300000} \approx \frac{150045.05}{1300000} \approx 0.1154 , \text{m}^3 \]
Now we can find \( n \):
Using the relation for polytropic expansion:
\[ \frac{P_1}{P_2} = \left( \frac{V_2}{V_1} \right)^n \]
Substitute the known values:
\[ \frac{1300000}{44559.525} = \left( \frac{2}{0.1154} \right)^n \]
Calculating the left side:
\[ \frac{1300000}{44559.525} \approx 29.17 \]
Calculating the right side:
\[ \frac{2}{0.1154} \approx 17.31 \]
Therefore,
\[ 29.17 = (17.31)^n \]
Taking the logarithm of both sides:
\[ \log(29.17) = n \log(17.31) \]
Calculating:
\[ \log(29.17) \approx 1.466 \quad \text{and} \quad \log(17.31) \approx 1.237 \]
Thus,
\[ n = \frac{1.466}{1.237} \approx 1.187 \]
2.4 Work Done During Expansion
The work done in a polytropic process is given by:
\[ W = \frac{P_2 V_2 - P_1 V_1}{n - 1} \]
Now, substituting the values we calculated:
\[ W = \frac{(44559.525 , \text{Pa}) (2 , \text{m}^3) - (1300000 , \text{Pa}) (0.1154 , \text{m}^3)}{1.187 - 1} \]
Calculating:
\[ W = \frac{89119.05 - 150045.0}{0.187} \]
Calculating:
\[ W \approx \frac{-60825.95}{0.187} \approx -325,054.4 , \text{J} \approx -325.05 , \text{kJ} \]
Summary of results:
- 2.1 Final Pressure after Expansion: \( P_2 \approx 44.56 , \text{kPa} \)
- 2.2 Index of Expansion: \( n \approx 1.187 \)
- 2.3 Initial Volume: \( V_1 \approx 0.1154 , \text{m}^3 \)
- 2.4 Work Done during Expansion: \( W \approx -325.05 , \text{kJ} \)
The work done is indicated as negative, which means that work has been done by the system during expansion.