At [tex]338 \, \text{mm Hg}[/tex] and [tex]72^{\circ} \text{C}[/tex], a sample of carbon monoxide gas occupies a volume of [tex]0.225 \, \text{L}[/tex]. The gas is transferred to a [tex]1.50 \, \text{L}[/tex] flask and the temperature is reduced to [tex]-15^{\circ} \text{C}[/tex]. What is the pressure of the gas in the flask?

a. [tex]8.91 \, \text{mm Hg}[/tex]
b. [tex]37.9 \, \text{mm Hg}[/tex]
c. [tex]67.8 \, \text{mm Hg}[/tex]
d. [tex]3.018 \times 10^3 \, \text{mm Hg}[/tex]
e. [tex]5.48 \times 10^4 \, \text{mm Hg}[/tex]



Answer :

To determine the pressure of the gas after the change in volume and temperature, we can use the combined gas law:

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

Here:
- \( P_1 \) = initial pressure = \( 338 \, \text{mm Hg} \)
- \( V_1 \) = initial volume = \( 0.225 \, \text{L} \)
- \( T_1 \) = initial temperature in Kelvin = \( 72^\circ \text{C} \) converted to Kelvin
- \( V_2 \) = final volume = \( 1.50 \, \text{L} \)
- \( T_2 \) = final temperature in Kelvin = \( -15^\circ \text{C} \) converted to Kelvin

First, let's convert the temperatures from Celsius to Kelvin:
[tex]\[ T_1 = 72 + 273.15 = 345.15 \, \text{K} \][/tex]
[tex]\[ T_2 = -15 + 273.15 = 258.15 \, \text{K} \][/tex]

Next, we'll rearrange the combined gas law to solve for the final pressure \( P_2 \):

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

Plugging in the given values:

[tex]\[ P_2 = \frac{338 \, \text{mm Hg} \times 0.225 \, \text{L} \times 258.15 \, \text{K}}{1.50 \, \text{L} \times 345.15 \, \text{K}} \][/tex]

Now, let's simplify and calculate:

[tex]\[ P_2 = \frac{19,669.3875}{517.725} \approx 37.92 \, \text{mm Hg} \][/tex]

Thus, the pressure of the gas in the flask is approximately \( 37.9 \, \text{mm Hg} \).

Therefore, the correct answer is:
b. [tex]\( 37.9 \, \text{mm Hg} \)[/tex]