Solution 15: Force Exerted by Gas
Step 1: Calculate the amount of Argon
Initially, at very low temperature, the argon exerts pressure $p_0$ only at the bottom, meaning it is in a condensed state. This pressure is purely due to the weight of the argon spread over the base area $A$. $$F_{bottom} = p_0 \times A$$ Since $F_{bottom} = \text{Weight} = nM g$ (where $n$ is moles, $M$ is molar mass), $$nM g = p_0 A \implies n = \frac{p_0 A}{Mg}$$ For a cube of volume $V$, the area $A = V^{2/3}$. $$n = \frac{p_0 V^{2/3}}{Mg}$$
Initially, at very low temperature, the argon exerts pressure $p_0$ only at the bottom, meaning it is in a condensed state. This pressure is purely due to the weight of the argon spread over the base area $A$. $$F_{bottom} = p_0 \times A$$ Since $F_{bottom} = \text{Weight} = nM g$ (where $n$ is moles, $M$ is molar mass), $$nM g = p_0 A \implies n = \frac{p_0 A}{Mg}$$ For a cube of volume $V$, the area $A = V^{2/3}$. $$n = \frac{p_0 V^{2/3}}{Mg}$$
Step 2: Gaseous State at Temperature T
When heated to temperature $T$, argon becomes a gas filling the volume $V$. Using the Ideal Gas Law $PV = nRT$: $$P = \frac{nRT}{V}$$ Substituting $n$: $$P = \frac{p_0 V^{2/3}}{Mg} \cdot \frac{RT}{V} = \frac{p_0 RT}{Mg V^{1/3}}$$
When heated to temperature $T$, argon becomes a gas filling the volume $V$. Using the Ideal Gas Law $PV = nRT$: $$P = \frac{nRT}{V}$$ Substituting $n$: $$P = \frac{p_0 V^{2/3}}{Mg} \cdot \frac{RT}{V} = \frac{p_0 RT}{Mg V^{1/3}}$$
Step 3: Force on Top Surface
The force on the top surface is due to gas pressure: $F = P \times A$. $$F = \left( \frac{p_0 RT}{Mg V^{1/3}} \right) \times V^{2/3}$$ $$F = \frac{p_0 V^{1/3} RT}{Mg}$$
The force on the top surface is due to gas pressure: $F = P \times A$. $$F = \left( \frac{p_0 RT}{Mg V^{1/3}} \right) \times V^{2/3}$$ $$F = \frac{p_0 V^{1/3} RT}{Mg}$$
Answer: $$F = \frac{p_0 V^{1/3} RT}{Mg}$$