Consider moist air with absolute humidity of 0.02 (kg moisture)/(kg dry air) at 1 bar pressure. The vapor pressure of water is given by the equation: \[ \ln P_{{sat}} = 12 - \frac{4000}{T - 40} \] where \( P_{{sat}} \) is in bar and \( T \) is in K. The molecular weight of water and dry air are 18 kg/kmol and 29 kg/kmol, respectively. The dew temperature of the moist air is ____________ ℃ (rounded off to the nearest integer).
\[ \ln P_{\text{sat}} = 12 - \frac{4000}{T - 40} \]
Step 2: Calculate the partial pressure of the water vapor:\[ y_{H_2O} = \frac{\text{absolute humidity} \times 1000}{M_{\text{water}}} \times \frac{M_{\text{air}}}{1000} \] \[ y_{H_2O} = \frac{0.02 \times 1000}{18} \times \frac{29}{1000} = 0.0322 \]
The partial pressure of water vapor is:
\[ P_{H_2O} = y_{H_2O} \times P_{\text{total}} = 0.0322 \times 1 = 0.0322 \, \text{bar} \]
Step 3: Solve for the dew temperature \( T \):\[ P_{\text{sat}} = P_{H_2O} = 0.0322 \, \text{bar} \] \[ \ln 0.0322 = 12 - \frac{4000}{T - 40} \] \[ -3.442 = 12 - \frac{4000}{T - 40} \] \[ \frac{4000}{T - 40} = 15.442 \Rightarrow T - 40 = \frac{4000}{15.442} = 259.4 \Rightarrow T = 259.4 + 40 = 299.4 \, K \] \[ T_{\text{dew}} = 299.4 - 273.15 = 26.25^\circ C \]
Therefore, the dew temperature is approximately 26°C.
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
The residence-time distribution (RTD) function of a reactor (in min$^{-1}$) is 
The mean residence time of the reactor is __________ min (rounded off to 2 decimal places).}