At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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
[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
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.