Find the pressure of air in a vessel being evacuated as a function of evacuation time `t`. The vessel volume is `V`, the initial pressure is `p_0`. The process is assumed to be isothermal, and the evacuation rate equal to `C` and independent of pressure.
The evatuation rate is the gas volume being evacuated per unit time, with that volume being measured under the gas pressure attained by that moment.

Explanation

To find the pressure of air in a vessel being evacuated as a function of evacuation time \( t \), we will follow these steps:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
where:
- \( P \) = pressure,
- \( V \) = volume,
- \( n \) = number of moles,
- \( R \) = universal gas constant,
- \( T \) = temperature.

Since the process is isothermal, \( T \) is constant.

### Step 2: Relate Moles to Mass
We can express the number of moles \( n \) in terms of mass \( m \) and molar mass \( M \):
\[ n = \frac{m}{M} \]
Substituting this into the ideal gas law gives:
\[ PV = \frac{m}{M}RT \]

### Step 3: Differentiate the Equation
To find how pressure changes with time, we differentiate the equation with respect to time \( t \):
\[ \frac{dP}{dt} V = \frac{RT}{M} \frac{dm}{dt} \]
Here, \( V \), \( R \), and \( T \) are constants.

### Step 4: Define the Evacuation Rate
The evacuation rate \( C \) is defined as the volume of gas being evacuated per unit time. We can express this as:
\[ C = -\frac{dV}{dt} \]
The volume of gas evacuated relates to the change in mass:
\[ C = -\frac{V}{M} \frac{dm}{dt} \]

### Step 5: Substitute into the Differential Equation
Substituting the expression for \( \frac{dm}{dt} \) into the differentiated ideal gas law gives:
\[ \frac{dP}{dt} = -\frac{C}{V} P \]

### Step 6: Solve the Differential Equation
This is a separable differential equation. Rearranging gives:
\[ \frac{dP}{P} = -\frac{C}{V} dt \]
Integrating both sides:
\[ \int \frac{dP}{P} = -\frac{C}{V} \int dt \]
This results in:
\[ \ln P = -\frac{C}{V} t + \ln P_0 \]
where \( P_0 \) is the initial pressure at \( t = 0 \).

### Step 7: Exponentiate to Solve for Pressure
Exponentiating both sides gives:
\[ P = P_0 e^{-\frac{C}{V} t} \]

### Final Result
Thus, the pressure of air in the vessel as a function of evacuation time \( t \) is:
\[ P(t) = P_0 e^{-\frac{C}{V} t} \]