Question

An ideal gas is kept in a closed container. If the temperature is doubled and the volume of the container is reduced to half, the gas pressure is:

• A. Unchanged
• B. Halved
• C. Doubled
• D. Increased by 4 times
• E. Increased by 16 times

Answer 1

The Correct Option is D

Given:

• Ideal gas in a closed container (fixed amount of gas, n = constant)
• Initial conditions: Temperature \( T_1 \), Volume \( V_1 \), Pressure \( P_1 \)
• Final conditions:
  • Temperature \( T_2 = 2T_1 \) (doubled)
  • Volume \( V_2 = \frac{V_1}{2} \) (halved)

Step 1: Apply Ideal Gas Law

The ideal gas law states:

\[ PV = nRT \]

For the initial and final states:

\[ P_1 V_1 = nRT_1 \]

\[ P_2 V_2 = nRT_2 \]

Step 2: Relate Pressures

Divide the final state equation by the initial state equation:

\[ \frac{P_2 V_2}{P_1 V_1} = \frac{T_2}{T_1} \]

Substitute \( V_2 = \frac{V_1}{2} \) and \( T_2 = 2T_1 \):

\[ \frac{P_2 \left(\frac{V_1}{2}\right)}{P_1 V_1} = \frac{2T_1}{T_1} \]

Simplify:

\[ \frac{P_2}{2P_1} = 2 \]

\[ P_2 = 4P_1 \]

Conclusion:

The gas pressure is increased by 4 times.

Answer: \(\boxed{D}\)

Answer 2

1. State the ideal gas law:

The ideal gas law relates pressure (P), volume (V), and temperature (T) of an ideal gas:

\[PV = nRT\]

where:

• n is the number of moles of gas (constant in this case)
• R is the ideal gas constant (constant)

2. Set up the initial and final conditions:

Initial conditions:

\[P_1V_1 = nRT_1\]

Final conditions (temperature is doubled, volume is halved):

\[P_2V_2 = nRT_2\]

\[P_2(\frac{1}{2}V_1) = nR(2T_1)\]

3. Solve for the final pressure:

Divide the final condition equation by the initial condition equation:

\[\frac{P_2(\frac{1}{2}V_1)}{P_1V_1} = \frac{nR(2T_1)}{nRT_1}\]

Simplify:

\[\frac{\frac{1}{2}P_2}{P_1} = \frac{2}{1}\]

\[\frac{P_2}{2P_1} = 2\]

\[P_2 = 4P_1\]

Therefore, the final pressure \(P_2\) is four times the initial pressure \(P_1\).