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: Concept
Use the Ideal Gas Equation in the form: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.

Step 2: Meaning
$P_2 = 2P_1$ and $T_2 = T_1/2$. We need to find $V_2$.

Step 3: Analysis
$V_2 = V_1 (\frac{P_1}{P_2}) (\frac{T_2}{T_1}) = V_1 (\frac{P_1}{2P_1}) (\frac{T_1/2}{T_1}) = V_1 (\frac{1}{2}) (\frac{1}{2})$.

Step 4: Conclusion
$V_2 = V_1/4$. The volume becomes one-fourth.
Final Answer: (A)