An amount n (in moles) of a monatomic gas at initial temperature `T_(0)` is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature `T_(s)(gtT_(0))` . And the atmospheric pressure is `P_(a)` . Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area A, thickness x and thermal conductivity K. Assuming all change to be slow, find the distance moved by the piston in time t.

Explanation

To solve the problem step by step, we will analyze the heat transfer and the behavior of the gas in the cylindrical vessel with a piston.

### Step 1: Understand the System
We have a monatomic gas in a cylinder with:
- Initial temperature \( T_0 \)
- Surrounding temperature \( T_s \) (where \( T_s > T_0 \))
- Atmospheric pressure \( P_a \)
- Surface area \( A \)
- Thickness \( x \)
- Thermal conductivity \( K \)
- Amount of gas \( n \) (in moles)

### Step 2: Heat Transfer Calculation
The rate of heat transfer \( \frac{dq}{dt} \) through the bottom of the cylinder can be calculated using Fourier's law of heat conduction:
\[ \frac{dq}{dt} = -\frac{K A (T_s - T_0)}{x} \]
This equation indicates that heat flows from the surrounding air into the gas.

### Step 3: Relate Heat Transfer to Temperature Change
The heat absorbed by the gas can be expressed as:
\[ dq = n C_p dT \]
For a monatomic gas, the specific heat at constant pressure \( C_p \) is given by:
\[ C_p = \frac{5}{2} R \]
Thus, we can write:
\[ dq = n \left(\frac{5}{2} R\right) dT \]

### Step 4: Set Up the Differential Equation
From the heat transfer and heat absorbed, we can equate:
\[ n \left(\frac{5}{2} R\right) dT = -\frac{K A (T_s - T_0)}{x} dt \]
Rearranging gives:
\[ \frac{dT}{T_s - T_0} = -\frac{2 K A}{5 n R x} dt \]

### Step 5: Integrate the Equation
Integrate both sides:
\[ \int_{T_0}^{T} \frac{dT}{T_s - T_0} = -\frac{2 K A}{5 n R x} \int_{0}^{t} dt \]
This leads to:
\[ \ln\left(\frac{T_s - T}{T_s - T_0}\right) = -\frac{2 K A t}{5 n R x} \]

### Step 6: Solve for Temperature
Exponentiating both sides, we have:
\[ \frac{T_s - T}{T_s - T_0} = e^{-\frac{2 K A t}{5 n R x}} \]
Rearranging gives:
\[ T = T_s - (T_s - T_0)e^{-\frac{2 K A t}{5 n R x}} \]

### Step 7: Relate Temperature Change to Piston Displacement
Using the ideal gas law \( PV = nRT \), we differentiate to find the relationship between the change in volume and temperature:
\[ P dV = n R dT \]
The change in volume \( dV \) can be expressed as:
\[ dV = A dx \]
Thus:
\[ P A dx = n R dT \]
From this, we can express \( dT \):
\[ dT = \frac{P A dx}{n R} \]

### Step 8: Substitute \( dT \) into the Temperature Equation
Substituting \( dT \) into our earlier equation gives:
\[ \frac{P A dx}{n R} = (T_s - T_0)e^{-\frac{2 K A t}{5 n R x}} \]
Rearranging this, we find:
\[ dx = \frac{n R}{P A} (T_s - T_0) (1 - e^{-\frac{2 K A t}{5 n R x}}) \]

### Step 9: Final Expression for Piston Displacement
Integrating this expression over time will yield the total displacement \( x \) of the piston.

### Final Result
The distance moved by the piston in time \( t \) is given by:
\[ x = \frac{n R}{P_a A} (T_s - T_0) \left(1 - e^{-\frac{2 K A t}{5 n R x}}\right) \]