A vertical cylinder is placed with a piston inside. The cross-sectional area of the piston is $S = 64 \text{ cm}^2$ and its mass is negligible. Inside the cylinder, there is air at normal atmospheric pressure and temperature $t = 19.9^\circ \text{C}$. The initial volume is $V_1 = 5.6 \text{ l}$. What mass of weight must be placed on the piston so that the volume of the air becomes $V_2 = 3 \text{ l}$? The molar mass of air is given as $M_0 = 29 \times 10^{-3} \text{ kg/mol}$.

1. Problem Description

A vertical cylinder is placed with a piston inside. The cross-sectional area of the piston is $S = 64 \text{ cm}^2$ and its mass is negligible. Inside the cylinder, there is air at normal atmospheric pressure and temperature $t = 19.9^\circ \text{C}$. The initial volume is $V_1 = 5.6 \text{ l}$. What mass of weight must be placed on the piston so that the volume of the air becomes $V_2 = 3 \text{ l}$? The molar mass of air is given as $M_0 = 29 \times 10^{-3} \text{ kg/mol}$.

2. Solution Steps

First, we convert all units to SI units.

$S = 64 \text{ cm}^2 = 64 \times 10^{-4} \text{ m}^2 = 0.0064 \text{ m}^2$

$V_1 = 5.6 \text{ l} = 5.6 \times 10^{-3} \text{ m}^3$

$V_2 = 3 \text{ l} = 3 \times 10^{-3} \text{ m}^3$

$t = 19.9^\circ \text{C} = 19.9 + 273.15 = 293.05 \text{ K}$

Since the process is assumed to be isothermal (constant temperature), we can apply Boyle's Law. After placing the mass, the pressure inside the cylinder will increase. The additional pressure is due to the weight of the added mass.

$P_1 V_1 = P_2 V_2$

$P_1$ is atmospheric pressure, $P_1 = 101325 \text{ Pa}$.

$P_2 = P_1 + \frac{mg}{S}$

Substituting $P_2$ into Boyle's Law gives:

$P_1 V_1 = (P_1 + \frac{mg}{S}) V_2$

$P_1 V_1 = P_1 V_2 + \frac{mg}{S} V_2$

$P_1 V_1 - P_1 V_2 = \frac{mg}{S} V_2$

$P_1 (V_1 - V_2) = \frac{mg}{S} V_2$

$m = \frac{P_1 S (V_1 - V_2)}{g V_2}$

$m = \frac{101325 \text{ Pa} \times 0.0064 \text{ m}^2 \times (5.6 \times 10^{-3} \text{ m}^3 - 3 \times 10^{-3} \text{ m}^3)}{9.81 \text{ m/s}^2 \times 3 \times 10^{-3} \text{ m}^3}$

$m = \frac{101325 \times 0.0064 \times 2.6 \times 10^{-3}}{9.81 \times 3 \times 10^{-3}}$

$m = \frac{1.684}{0.02943} \times 10^{-3} \times 10^3$

$m \approx 57.22 \text{ kg} \times \frac{0.0026}{0.003} \approx \frac{1.68388}{0.02943} \approx 57.21 \text{ kg}$

3. Final Answer

The mass of the weight that must be placed on the piston is approximately 57.21 kg.