Question

The temperature of an ideal gas increases \(2\) times when compressed adiabatically to half the volume. Its equation of state is

• A. \(PV^{3/2}=\text{constant}\)
• B. \(PV^{4/3}=\text{constant}\)
• C. \(PV^{1/2}=\text{constant}\)
• D. \(PV^{2}=\text{constant}\)

Hint:

Use \(TV^{\gamma-1}=\) constant to find \(\gamma\), then write \(PV^{\gamma}=\) constant.

Answer 1

The Correct Option is D

Step 1: Use Gas Law with Adiabatic Relation
Combine $P_{1}V_{1}^{\gamma}=P_{2}V_{2}^{\gamma}$ with $\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$.

Step 2: Substitute
Pressure becomes $P_{2}=4P_{1}$ from the gas law ($2\times 2$), and $4=2^{\gamma}$.

Step 3: Result
$\gamma=2$, so $PV^{2}=$ constant, option (D).

Final Answer:
A pressure rise of four times for half the volume gives gamma equal to 2. \[ \boxed{\text{(D) }PV^{2}=\text{constant}} \]