According to the ideal gas law, by what factor does the temperature of a gas change if both its volume and pressure are doubled, assuming the amount of gas remains constant?

• A. The temperature increases by a factor of 2.
• B. The temperature remains unchanged.
• C. The temperature decreases by a factor of 2.
• D. The temperature increases by a factor of 4.

Answer: (D)

Verified step by step guidance

1. Recall the ideal gas law: \(P \times V = n \times R \times T\), where \(P\) is pressure, \(V\) is volume, \(n\) is the amount of gas (in moles), \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin.
2. Since the amount of gas \(n\) and the gas constant \(R\) remain constant, we can write the relationship as \(P \times V \propto T\).
3. Initially, let the pressure, volume, and temperature be \(P_1\), \(V_1\), and \(T_1\), respectively. Then, \(P_1 \times V_1 = n \times R \times T_1\).
4. If both pressure and volume are doubled, the new pressure and volume are \(P_2 = 2P_1\) and \(V_2 = 2V_1\). Substitute these into the ideal gas law: \(P_2 \times V_2 = n \times R \times T_2\).
5. Calculate the factor by which temperature changes by comparing \(T_2\) to \(T_1\): \(T_2 = \frac{P_2 \times V_2}{n \times R} = \frac{2P_1 \times 2V_1}{n \times R} = 4 \times \frac{P_1 \times V_1}{n \times R} = 4T_1\). Thus, the temperature increases by a factor of 4.