Certainly! Let's go through the steps to solve the problem using the relationship between pressure and temperature for a gas at constant volume, known as Gay-Lussac's Law.
According to Gay-Lussac's Law, for an ideal gas at a constant volume, the pressure of the gas is directly proportional to its absolute temperature. This relationship can be expressed as:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]
Given:
- Initial pressure, [tex]\(P_1 = 0.82 \, \text{atm}\)[/tex]
- Initial temperature, [tex]\(T_1 = 21^{\circ} \text{C}\)[/tex]
- Final temperature, [tex]\(T_2 = -3.5^{\circ} \text{C}\)[/tex]
First, we need to convert the temperatures from Celsius to Kelvin because the absolute temperature in these calculations must be in Kelvin:
[tex]\[ T_K = T_C + 273.15 \][/tex]
So for the initial temperature:
[tex]\[ T_1 = 21 + 273.15 = 294.15 \, \text{K} \][/tex]
And for the final temperature:
[tex]\[ T_2 = -3.5 + 273.15 = 269.65 \, \text{K} \][/tex]
Now we can apply the formula:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]
We need to solve for [tex]\(P_2\)[/tex]. Rearrange the equation to:
[tex]\[ P_2 = P_1 \frac{T_2}{T_1} \][/tex]
Substitute the known values:
[tex]\[ P_2 = 0.82 \times \frac{269.65}{294.15} \][/tex]
Perform the calculation:
[tex]\[ P_2 = 0.82 \times 0.9163 \][/tex]
[tex]\[ P_2 \approx 0.7517 \, \text{atm} \][/tex]
Therefore, the pressure after the temperature decrease is approximately [tex]\(0.7517 \, \text{atm}\)[/tex].
Given the choices, the closest value is:
[tex]\[ 0.75 \, \text{atm} \][/tex]
So, the correct answer is [tex]\( 0.75 \, \text{atm} \)[/tex].
