Physics: Thermal Physics - Boyle's Law, Charles's Law, and Gas Laws

A gas occupies a volume of $400\,cm^3$ at a pressure of $120\,kPa$. If the volume is increased to $600\,cm^3$ while the temperature remains constant, calculate the new pressure.

Given: $P_1 = 120\,kPa$, $V_1 = 400\,cm^3$, $V_2 = 600\,cm^3$. \nSince temperature is constant, we use Boyle's Law: $P_1 V_1 = P_2 V_2$ $120 \times 400 = P_2 \times 600$ $48000 = 600 P_2$ $P_2 = \frac{48000}{600} = 80\,kPa$

Because the volume increased, the particles have more space to move, resulting in fewer collisions with the walls. Therefore, the pressure decreased from $120\,kPa$ to $80\,kPa$.

A sample of gas has a volume of $2.5\,L$ at $27^{\circ}C$. Calculate the volume of the gas if the temperature is raised to $77^{\circ}C$ at a constant pressure.

Step 1: Convert temperatures to Kelvin: $T_1 = 27 + 273 = 300\,K$ $T_2 = 77 + 273 = 350\,K$ \nStep 2: Apply Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ $\frac{2.5}{300} = \frac{V_2}{350}$ $V_2 = \frac{2.5 \times 350}{300}$ $V_2 = \frac{875}{300} \approx 2.92\,L$

According to Charles's Law, volume is directly proportional to absolute temperature. As the temperature in Kelvin increases, the gas particles move faster and push the container walls further out to maintain constant pressure.

A rigid cylinder contains gas at a pressure of $200\,kPa$ at a temperature of $300\,K$. If the cylinder is heated until the pressure reaches $300\,kPa$, what is the final temperature?

Given: $P_1 = 200\,kPa$, $T_1 = 300\,K$, $P_2 = 300\,kPa$. \nUsing Gay-Lussac's Law (since the cylinder is rigid, volume $V$ is constant): $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ $\frac{200}{300} = \frac{300}{T_2}$ $200 \times T_2 = 300 \times 300$ $200 T_2 = 90000$ $T_2 = \frac{90000}{200} = 450\,K$

In a fixed volume, increasing the temperature increases the kinetic energy of the particles, leading to more frequent and more forceful collisions, which increases the pressure.