Question

the volume of a quantity of gas decreases by 50% when it cools from 20 °c to 10 °c. does the pressure of the gas increase, decrease or remain the same?
○ increase
○ decrease
○ reamains the same

Explanation:

Step1: Recall Ideal Gas Law

The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles (constant here), \( R \) is gas constant, and \( T \) is absolute temperature (in Kelvin).

Step2: Convert Temperatures to Kelvin

Initial temperature \( T_1 = 20^\circ\text{C} + 273.15 = 293.15\,\text{K} \).
Final temperature \( T_2 = 10^\circ\text{C} + 273.15 = 283.15\,\text{K} \).

Step3: Analyze Volume Change

Volume decreases by 50%, so \( V_2 = 0.5V_1 \).

Step4: Apply Ideal Gas Law for Initial and Final States

For initial state: \( P_1V_1 = nRT_1 \).
For final state: \( P_2V_2 = nRT_2 \).

Divide the final equation by the initial equation:
\( \frac{P_2V_2}{P_1V_1} = \frac{T_2}{T_1} \).

Substitute \( V_2 = 0.5V_1 \), \( T_1 = 293.15\,\text{K} \), \( T_2 = 283.15\,\text{K} \):
\( \frac{P_2(0.5V_1)}{P_1V_1} = \frac{283.15}{293.15} \).

Simplify: \( \frac{0.5P_2}{P_1} \approx 0.966 \).

Solve for \( \frac{P_2}{P_1} \): \( P_2 \approx 0.966 \times 2 P_1 \approx 1.932P_1 \).

Since \( P_2 > P_1 \), pressure increases.

Answer:

A. increase