If the pressure of the gas is doubled, keeping its temperature constant, the volume of the gas is:

To analyze the relationship between pressure, volume, and temperature for an ideal gas, we'll apply Boyle's Law, which states that:

$PV = k$

where:

• $P$ = pressure of the gas
• $V$ = volume of the gas
• $k$ = constant

1. ### Break Down the Problem

• We are given that the pressure of the gas is doubled while keeping the temperature constant.
• We need to find out how the volume of the gas changes as a result.

2. ### Relevant Concepts

According to Boyle's Law: $P_1V_1 = P_2V_2$

Where:

• $P_1$ and $V_1$ are the initial pressure and volume.
• $P_2$ and $V_2$ are the final pressure and volume.

3. ### Analysis and Detail

Let:

• The initial pressure $P_1 = P$ and volume $V_1 = V$.
• The final pressure $P_2 = 2P$ (since it's doubled).

According to Boyle's Law: $PV = (2P)V_2$

Now we can solve for $V_2$: $PV = 2PV_2$

Dividing both sides by $2P$ (assuming $P \neq 0$): $V_2 = \frac{V}{2}$

4. ### Verify and Summarize

The relationship derived shows that when the pressure is doubled while keeping the temperature constant, the volume of the gas is halved.

Final Answer

Thus, if the pressure of the gas is doubled, keeping its temperature constant, the volume of the gas is: $V_2 = \frac{V}{2}$