Let's start solving the problem step-by-step.
### Problem:
If the initial temperature of an ideal gas at 2.250 atm is 62.00°C, what final temperature would cause the pressure to be reduced to 1.650 atm?
### Given Data:
- Initial pressure ([tex]\(P_1\)[/tex]) = 2.250 atm
- Initial temperature ([tex]\(T_1\)[/tex]) = 62.00°C
- Final pressure ([tex]\(P_2\)[/tex]) = 1.650 atm
### Step 1: Convert the Initial Temperature to Kelvin
First, we need to convert the initial temperature from Celsius to Kelvin because gas laws typically require temperatures in Kelvin.
[tex]\[ T(\text{K}) = T(\text{°C}) + 273.15 \][/tex]
So,
[tex]\[ T_1 (\text{K}) = 62.00 + 273.15 = 335.15 \text{ K} \][/tex]
### Step 2: Use the Ideal Gas Law (Directly Proportional Relationship)
For an ideal gas under constant volume conditions, the pressure and temperature are directly proportional. This relationship is given by the equation:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]
Where:
- [tex]\(P_1\)[/tex] and [tex]\(T_1\)[/tex] are the initial pressure and temperature,
- [tex]\(P_2\)[/tex] and [tex]\(T_2\)[/tex] are the final pressure and temperature.
Rearranging this equation to solve for the final temperature ([tex]\(T_2\)[/tex]):
[tex]\[ T_2 = T_1 \cdot \frac{P_2}{P_1} \][/tex]
### Step 3: Plug in the Known Values
[tex]\[ T_2 = 335.15 \, \text{K} \cdot \frac{1.650 \, \text{atm}}{2.250 \, \text{atm}} \][/tex]
[tex]\[ T_2 = 335.15 \cdot \frac{1.650}{2.250} \][/tex]
[tex]\[ T_2 \approx 245.7767 \text{ K} \][/tex]
### Step 4: Convert the Final Temperature Back to Celsius
Now, convert the final temperature from Kelvin back to Celsius:
[tex]\[ T(\text{°C}) = T(\text{K}) - 273.15 \][/tex]
So,
[tex]\[ T_2 (\text{°C}) = 245.7767 - 273.15 \approx -27.3733 \text{ °C} \][/tex]
### Conclusion:
The final temperature that would cause the pressure to be reduced to 1.650 atm is approximately [tex]\( -27.37 \, \text{°C} \)[/tex].
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Thus, the results step-by-step are:
- Initial temperature in Kelvin: 335.15 K
- Final temperature in Kelvin: 245.7767 K
- Final temperature in Celsius: -27.3733°C
