Question

If the pressure of an ideal gas is doubled and its absolute temperature is halved; the volume will become:

• A. $1/4$ of initial volume
• B. $1/2$ initial volume
• C. Same as initial volume
• D. 2 times of initial volume

Hint:

Squeezing it harder (Pressure $\times$ 2) makes it smaller, and cooling it down (Temp 2) makes it even smaller. Both effects multiply!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the change in volume, we use the Combined Gas Law, which relates pressure, volume, and temperature for a fixed amount of an ideal gas.
Step 2: Key Formula or Approach:
Ideal Gas Law: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)
We need to find \( V_2 \) in terms of \( V_1 \).
Step 3: Detailed Explanation:
Given conditions: * \( P_2 = 2P_1 \) (Pressure doubled) * \( T_2 = \frac{1}{2}T_1 \) (Temperature halved) Substitute these into the equation: \[ \frac{P_1 V_1}{T_1} = \frac{(2P_1) V_2}{(T_1/2)} \] Multiply both sides by \( T_1 \) and divide by \( P_1 \): \[ V_1 = \frac{2 V_2}{1/2} \] \[ V_1 = 4V_2 \] \[ V_2 = \frac{1}{4}V_1 \]
Step 4: Final Answer:
The volume will become 1/4 of the initial volume.