When the pressure on 200 cm3 of a gas is doubled, and its absolute temperature is halved, its volume would become

Here's the solution for question 6:

6. When the pressure on $200 \text{ cm}^3$ of a gas is doubled, and its absolute temperature is halved, its volume would become

• Step 1: State the combined gas law formula. The combined gas law relates pressure, volume, and absolute temperature for a fixed amount of gas: $$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$ where $P$ is pressure, $V$ is volume, and $T$ is absolute temperature.
• Step 2: Identify the initial and final conditions from the problem statement.
  • Initial volume $V_1 = 200 \text{ cm}^3$.
  • Let initial pressure $P_1 = P$.
  • Let initial absolute temperature $T_1 = T$.
  • The pressure is doubled, so final pressure $P_2 = 2P$.
  • The absolute temperature is halved, so final absolute temperature $T_2 = \frac{T}{2}$.
  • We need to find the final volume $V_2$.
• Step 3: Substitute these values into the combined gas law equation. $$\frac{P \cdot 200 \text{ cm}^3}{T} = \frac{(2P) \cdot V_2}{(T/2)}$$
• Step 4: Simplify the equation to solve for $V_2$. First, simplify the right side of the equation: $$\frac{P \cdot 200 \text{ cm}^3}{T} = \frac{2P \cdot V_2 \cdot 2}{T}$$ $$\frac{P \cdot 200 \text{ cm}^3}{T} = \frac{4P \cdot V_2}{T}$$ Now, multiply both sides by $T$ and divide by $P$: $$200 \text{ cm}^3 = 4V_2$$ $$V_2 = \frac{200 \text{ cm}^3}{4}$$ $$V_2 = 50 \text{ cm}^3$$

The correct option is C.

• C) $50 \text{ cm}^3$