To solve the problem, we will go through each part step by step.
2.1 Mass of air in the cylinder
The mass of air can be calculated using the ideal gas equation:
\[ PV = mRT \]
Where:
- \( P \) = absolute pressure (Pa)
- \( V \) = volume (m³)
- \( m \) = mass (kg)
- \( R \) = specific gas constant for air (approximately \( 287 , \text{J/(kg·K)} \))
- \( T \) = absolute temperature (K)
First, we need to convert the volume from liters to cubic meters:
\[ V = 60 , \text{liters} = 60 \times 10^{-3} , \text{m}^3 = 0.06 , \text{m}^3 \]
Converting the temperature from Celsius to Kelvin:
\[ T = 27 °C + 273.15 = 300.15 , K \]
Now substituting the values into the ideal gas equation:
\[ 200 \times 10^3 , \text{Pa} \cdot 0.06 , \text{m}^3 = m \cdot 287 , \frac{\text{J}}{\text{kg·K}} \cdot 300.15 , K \]
Solving for mass \( m \):
\[ m = \frac{200 \times 10^3 \cdot 0.06}{287 \cdot 300.15} \] \[ m \approx \frac{12000}{86119.05} \approx 0.139 , \text{kg} \]
2.2 Final absolute temperature
Since the pressure is constant, we can use the ideal gas law again to find the final temperature with the new volume:
\[ P_1V_1/T_1 = P_2V_2/T_2 \]
Given that \( P_1 = P_2 = 200 , \text{kPa} = 200 \times 10^3 , \text{Pa} \), we can simplify the equation because pressure cancels out:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where:
- \( V_1 = 0.06 , \text{m}^3 \)
- \( V_2 = 240 , \text{liters} = 240 \times 10^{-3} , \text{m}^3 = 0.24 , \text{m}^3 \)
Substituting the known values:
\[ \frac{0.06}{300.15} = \frac{0.24}{T_2} \]
Solving for \( T_2 \):
\[ T_2 = \frac{0.24 \cdot 300.15}{0.06} \approx 1200.6 , K \]
2.3 Change in internal energy
The change in internal energy for an ideal gas can be calculated using the formula:
\[ \Delta U = m C_v \Delta T \]
Where:
- \( \Delta T = T_2 - T_1 \)
Now calculate \( \Delta T \):
\[ \Delta T = 1200.6 - 300.15 = 900.45 , K \]
Now substituting the known values into the internal energy formula:
\[ \Delta U = 0.139 \cdot 720 \cdot 900.45 \] \[ \Delta U \approx 0.139 \cdot 720 \cdot 900.45 \approx 90,049.5 , \text{J} \]
2.4 Work done
For a constant pressure process, work done is given by:
\[ W = P \Delta V \]
Where:
- \( \Delta V = V_2 - V_1 = 0.24 - 0.06 = 0.18 , \text{m}^3 \)
Now applying the values:
\[ W = 200 \times 10^3 \cdot 0.18 = 36,000 , \text{J} \]
2.5 Heat flow
Using the first law of thermodynamics:
\[ Q = \Delta U + W \]
Substituting the values of \( \Delta U \) and \( W \):
\[ Q = 90,049.5 + 36,000 \] \[ Q \approx 126,049.5 , \text{J} \]
Summary of Calculations:
- Mass of air: \( 0.139 , \text{kg}** \)
- Final absolute temperature: \( 1200.6 , K** \)
- Change in internal energy: \( 90,049.5 , \text{J}** \)
- Work done: \( 36,000 , \text{J}** \)
- Heat flow: \( 126,049.5 , \text{J}** \)