1. Kinetic Theory Analysis of Pressure

The pressure exerted by a gas on a wall is due to the change in momentum of the gas molecules striking and rebounding from the surface. In the kinetic theory of gases, the pressure ($P$) is directly proportional to the mean kinetic energy density of the molecules.

Since the kinetic energy corresponds to the temperature ($KE \propto T$), we can model the pressure contribution of a stream of molecules as being proportional to their temperature: $$ P \propto T $$

The ambient pressure $p_0$ exists when both incident and reflected molecules correspond to the ambient temperature $T_0$. We can view the total pressure on a surface as the sum of the partial pressure from the incident stream ($P_{in}$) and the partial pressure from the reflected stream ($P_{out}$).

• Incident Stream: Comes from the ambient air at $T_0$. It contributes half of the ambient pressure: $$ P_{in} = \frac{p_0}{2} $$
• Reflected Stream: Leaves the surface at the surface temperature $T_{surf}$. Its pressure contribution scales with the ratio of the surface temperature to the ambient temperature: $$ P_{out} = \frac{p_0}{2} \left( \frac{T_{surf}}{T_0} \right) $$

Therefore, the total pressure ($P_{total}$) on a surface at temperature $T_{surf}$ is: $$ P_{total} = P_{in} + P_{out} = \frac{p_0}{2} \left( 1 + \frac{T_{surf}}{T_0} \right) $$

2. Calculating Forces on the Sheet

Let the area of the sheet be $S$.

Top Surface (Temperature $T_1$):
The pressure on the top surface pushes downwards. $$ P_1 = \frac{p_0}{2} \left( 1 + \frac{T_1}{T_0} \right) $$ $$ F_1 = P_1 S = \frac{p_0 S}{2} \left( 1 + \frac{T_1}{T_0} \right) \quad (\text{downward}) $$

Bottom Surface (Temperature $T_2$):
The pressure on the bottom surface pushes upwards. $$ P_2 = \frac{p_0}{2} \left( 1 + \frac{T_2}{T_0} \right) $$ $$ F_2 = P_2 S = \frac{p_0 S}{2} \left( 1 + \frac{T_2}{T_0} \right) \quad (\text{upward}) $$

3. Equilibrium Condition

For the sheet to stay floating at a level height, the net vertical force must be zero. The upward force from the bottom air must balance the downward force from the top air and the gravitational force ($mg$).

$$ \Sigma F_y = 0 $$ $$ F_2 – F_1 – mg = 0 $$ $$ mg = F_2 – F_1 $$

4. Solving for Mass

Substitute the expressions for $F_1$ and $F_2$ into the equilibrium equation:

$$ mg = \left[ \frac{p_0 S}{2} \left( 1 + \frac{T_2}{T_0} \right) \right] – \left[ \frac{p_0 S}{2} \left( 1 + \frac{T_1}{T_0} \right) \right] $$

Factor out common terms:

$$ mg = \frac{p_0 S}{2} \left[ \left( 1 + \frac{T_2}{T_0} \right) – \left( 1 + \frac{T_1}{T_0} \right) \right] $$

Simplify the terms inside the bracket:

$$ mg = \frac{p_0 S}{2} \left[ 1 + \frac{T_2}{T_0} – 1 – \frac{T_1}{T_0} \right] $$ $$ mg = \frac{p_0 S}{2} \left[ \frac{T_2 – T_1}{T_0} \right] $$

Solve for $m$:

$$ m = \frac{S p_0 (T_2 – T_1)}{2 g T_0} $$