Questions: The temperature of a gas is 28°C, with a gas volume of 6.3 L, which exerts a pressure of 752 mm Hg. What will be the new temperature if the pressure decreases to 720 mm Hg, and the volume increased to 6.8 L?

Solution Steps

First, we need to convert the initial temperature from Celsius to Kelvin. The formula for this conversion is: \[ T(K) = T(^{\circ}C) + 273.15 \]

Given: \[ T_1 = 28^{\circ}C \]

So, \[ T_1(K) = 28 + 273.15 = 301.15 \, \text{K} \]

The combined gas law relates the pressure, volume, and temperature of a gas: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Given: \[ P_1 = 752 \, \text{mm Hg} \] \[ V_1 = 6.3 \, \text{L} \] \[ T_1 = 301.15 \, \text{K} \] \[ P_2 = 720 \, \text{mm Hg} \] \[ V_2 = 6.8 \, \text{L} \]

We need to find the new temperature \( T_2 \).

Rearrange the equation to solve for \( T_2 \): \[ T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} \]

Substitute the given values into the equation: \[ T_2 = \frac{720 \, \text{mm Hg} \times 6.8 \, \text{L} \times 301.15 \, \text{K}}{752 \, \text{mm Hg} \times 6.3 \, \text{L}} \]

Perform the calculation: \[ T_2 = \frac{720 \times 6.8 \times 301.15}{752 \times 6.3} \] \[ T_2 = \frac{147,556.56}{4,737.6} \] \[ T_2 \approx 31.14 \, \text{K} \]

Convert the new temperature from Kelvin back to Celsius: \[ T_2(^{\circ}C) = T_2(K) - 273.15 \] \[ T_2(^{\circ}C) = 31.14 - 273.15 \] \[ T_2(^{\circ}C) \approx -242.01 \]

Final Answer