Question

When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of gas

• A. will be half of original volume
• B. will be 4 times the original volume
• C. will be 2 times the original volume
• D. will be 1/4th times the original volume

Answer 1

The Correct Option is B

We can use the ideal gas law to solve this problem:

Ideal Gas Law: \( PV = nRT \)

Let the initial volume be \( V_1 \), pressure be \( P_1 \), and temperature be \( T_1 \). The initial conditions are:

\[ P_1 V_1 = n R T_1 \] After the changes: - The temperature is doubled: \( T_2 = 2T_1 \) - The pressure is halved: \( P_2 = \frac{P_1}{2} \) The new volume \( V_2 \) can be found using the ideal gas law: \[ P_2 V_2 = n R T_2 \] Substituting the changes: \[ \frac{P_1}{2} V_2 = n R (2T_1) \] Simplifying: \[ \frac{P_1}{2} V_2 = 2n R T_1 \] \[ V_2 = 4V_1 \] Thus, the volume of the gas will be 4 times the original volume when the temperature is doubled and the pressure is halved.

The correct answer is (B) : will be 4 times the original volume.

Answer 2

To solve this problem, we need to apply the ideal gas law, which relates pressure, volume, and temperature:

1. Ideal Gas Law:
The ideal gas law is given by:

$ PV = nRT $

Where:
- $P$ is the pressure of the gas
- $V$ is the volume of the gas
- $n$ is the number of moles of gas
- $R$ is the ideal gas constant
- $T$ is the absolute temperature of the gas.

2. Understanding the Changes in Conditions:
The problem states that the absolute temperature of the gas is doubled, and the pressure is halved. We need to find how these changes affect the volume of the gas. Let the initial temperature be $T_1$, the initial pressure be $P_1$, and the initial volume be $V_1$. After the changes, the new temperature is $T_2 = 2T_1$, and the new pressure is $P_2 = \frac{P_1}{2}$.

3. Setting Up the Ratio Using the Ideal Gas Law:
Using the ideal gas law for the initial and final states, we have:

$ P_1 V_1 = nRT_1 $ (initial state)
$ P_2 V_2 = nRT_2 $ (final state)

Now, substitute $P_2 = \frac{P_1}{2}$ and $T_2 = 2T_1$ into the second equation:

$ \frac{P_1}{2} V_2 = nR (2T_1) $
By simplifying, we get:

$ \frac{P_1}{2} V_2 = 2 nRT_1 $
Divide both sides by $P_1$ and compare the two equations:

$ \frac{1}{2} V_2 = 2 V_1 $
Multiplying both sides by 2:

$ V_2 = 4 V_1 $

Final Answer:
The volume of the gas will be 4 times the original volume. Therefore, the correct answer is (B) "4 times the original volume."