Let's solve the problem step-by-step using the Combined Gas Law, which relates the pressure, volume, and temperature of a gas:
[tex]\[ \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \][/tex]
where:
- [tex]\( P1 \)[/tex] = initial pressure
- [tex]\( V1 \)[/tex] = initial volume
- [tex]\( T1 \)[/tex] = initial temperature
- [tex]\( P2 \)[/tex] = final pressure
- [tex]\( V2 \)[/tex] = final volume
- [tex]\( T2 \)[/tex] = final temperature (which we need to find)
We are given the following values:
- [tex]\( P1 = 1.21 \)[/tex] atm
- [tex]\( V1 = 3.75 \)[/tex] L
- [tex]\( T1 = 293 \)[/tex] K
- [tex]\( P2 = 2.52 \)[/tex] atm
- [tex]\( V2 = 1.72 \)[/tex] L
We need to solve for [tex]\( T2 \)[/tex]:
Rearrange the Combined Gas Law to solve for [tex]\( T2 \)[/tex]:
[tex]\[ T2 = \frac{P2 \cdot V2 \cdot T1}{P1 \cdot V1} \][/tex]
Substitute the given values into the equation:
[tex]\[ T2 = \frac{2.52 \, \text{atm} \cdot 1.72 \, \text{L} \cdot 293 \, \text{K}}{1.21 \, \text{atm} \cdot 3.75 \, \text{L}} \][/tex]
Now, evaluate the numerator and the denominator separately:
Numerator:
[tex]\[ 2.52 \cdot 1.72 \cdot 293 = 1269.1936 \][/tex]
Denominator:
[tex]\[ 1.21 \cdot 3.75 = 4.5375 \][/tex]
Now, divide the results:
[tex]\[ T2 = \frac{1269.1936}{4.5375} \approx 279.885 \, \text{K} \][/tex]
The final temperature [tex]\( T2 \approx 280 \, \text{K} \)[/tex].
Therefore, the new temperature of the system is approximately:
C. 280 K
