Thermal Physics - Kinetic particle model of matter

A gas occupies a volume of $0.40\text{ m}^3$ at a pressure of $1.5 \times 10^5\text{ Pa}$. If the volume is compressed to $0.10\text{ m}^3$ at a constant temperature, calculate the new pressure.

$P_1 V_1 = P_2 V_2$ $(1.5 \times 10^5\text{ Pa}) \times (0.40\text{ m}^3) = P_2 \times (0.10\text{ m}^3)$ $P_2 = \frac{1.5 \times 10^5 \times 0.40}{0.10}$ $P_2 = 6.0 \times 10^5\text{ Pa}$

Since the temperature is constant, we apply Boyle's Law. Reducing the volume by a factor of 4 (from $0.40$ to $0.10$) results in the pressure increasing by a factor of 4.

Convert a room temperature of $25^\circ\text{C}$ into the Kelvin scale.

$T\text{ (K)} = \theta\text{ (}^\circ\text{C)} + 273$ $T = 25 + 273$ $T = 298\text{ K}$

To convert from Celsius to Kelvin, add 273 to the Celsius value. Kelvin is the absolute temperature scale used in gas law calculations.

Describe how the pressure of a gas in a rigid container changes if the temperature is increased.

As temperature increases, the average kinetic energy of the particles increases, meaning they move faster ($E_k = \frac{1}{2}mv^2$). This leads to more frequent collisions with the walls and more forceful collisions (greater change in momentum $\Delta p$). Consequently, the total force $F$ on the walls increases, leading to an increase in pressure $P$.

In a rigid container, volume is constant. According to the kinetic theory, pressure is directly proportional to the absolute temperature ($P \propto T$).