To find the final pressure and temperature of the mixture after the partition is removed, we can apply the principles of conservation of mass, energy, and the ideal gas law.
Parameters
Gas 1:
- Mass \( m_1 = 1 \text{ kg} \)
- Pressure \( P_1 = 500 \text{ kPa} \)
- Temperature \( T_1 = 300 \text{ K} \)
Gas 2:
- Mass \( m_2 = 1 \text{ kg} \)
- Pressure \( P_2 = 200 \text{ kPa} \)
- Temperature \( T_2 = 400 \text{ K} \)
Total Mass: \[ m_{\text{total}} = m_1 + m_2 = 1 \text{ kg} + 1 \text{ kg} = 2 \text{ kg} \]
Step 1: Calculate the number of moles of each gas
Using the ideal gas equation: \[ PV = nRT \] Where:
- \( P \) = pressure
- \( V \) = volume (not needed directly, but can be used if needed)
- \( n \) = number of moles
- \( R \) = specific gas constant, which we won't need here yet
- \( T \) = temperature
We can rearrange it to find the number of moles \( n \): \[ n = \frac{PV}{RT} \]
For both gases, we can use the relation to find their respective states as no volumes are provided, and we are considering the two gases in a rigid container.
Step 2: Calculate the overall temperature of the mixture
In an adiabatic process where no heat transfer occurs, the final temperature \( T_f \) can be found using the weighted average based on the internal energy balance of the two gases.
Assuming both gases can be treated ideally, and we are using specific heats, which we’ll assume each gas corresponds to a molar heat capacity \( C_v \).
- The internal energy of an ideal gas can be expressed as: \[ U = n C_v T \]
- The final temperature can be calculated from the energy conservation: \[ m_1 C_{v1} T_1 + m_2 C_{v2} T_2 = (m_1 + m_2) C_{v_f} T_f \]
For simplification, assume specific heats are equal \( C_{v1} = C_{v2} = C_v \): \[ m_1 T_1 + m_2 T_2 = (m_1 + m_2) T_f \] Plugging in the masses and temperatures: \[ 1 \text{ kg} \cdot 300 \text{ K} + 1 \text{ kg} \cdot 400 \text{ K} = 2 \text{ kg} \cdot T_f \] \[ 300 + 400 = 2 T_f \] \[ 700 = 2 T_f \implies T_f = 350 \text{ K} \]
Step 3: Calculate final pressure using Ideal Gas Law
To find the final pressure \( P_f \), we'll use the total number of moles found and the volume they occupy. However, we are working with relative values:
- Since the number of moles of each gas isn't quantified directly without additional parameters (like \( R \) for specific gases), we can derive \( P_f \) directly:
Knowing changing the volume will be unnecessary (rigid container), we’ll relate pressures and volumes: Using: \[ P_1 V + P_2 V = P_f V \]
- Assuming equal volumes, we’ll simplify: \[ \frac{P_f}{T_f} = \frac{P_1}{T_1} + \frac{P_2}{T_2} \] Let’s express each component:
- \( P_1 / T_1 = 500 / 300 \) kPa/K = \( \approx 1.66667 \)
- \( P_2 / T_2 = 200 / 400 \) kPa/K = \( 0.5 \)
Adding these: \[ P_f = \left(\frac{500 \text{ kPa}}{300 \text{ K}} + \frac{200 \text{ kPa}}{400 \text{ K}} \right) \cdot 350 \text{ K} \] Calculating: \[ \frac{500}{300} + \frac{200}{400} = 1.66667 + 0.5 = 2.16667 \] Now calculating: \[ P_f = 2.16667 \cdot 350 \approx 759.3345 \text{ kPa} \]
Final Results
- Final temperature \( T_f = 350 \text{ K} \)
- Final pressure \( P_f \approx 759.33 \text{ kPa} \)
This concludes our calculations for the final state of the mixture.