Given:
[tex]P_{1} = \text{25.0 atm} \times \dfrac{\text{760 mmHg}}{\text{1 atm}} = \text{19000 mmHg}[/tex]
[tex]T_{1} = \text{290 K}[/tex]
[tex]P_{2} = \text{1900 mmHg}[/tex]
Note: The initial pressure is in atm. We must convert it to mmHg to have the same unit with the final pressure which is expressed in mmHg.
Required:
[tex]T_{2}[/tex]
Strategy:
This is a gas law problem. What gas law should we use?
Since the given quantities are pressure and temperature, we will use Gay-Lussac's law. According to this gas law, the pressure of a gas is directly proportional to its absolute temperature keeping the volume and the amount of gas constant.
Caution: The temperature must be converted to kelvin. If the given temperature is in degree Celsius, add 273 or 273.15 to convert it to kelvin.
The formula used for Gay-Lussac's law is
[tex]\boxed{\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}}[/tex]
where:
[tex]P_{1} = \text{initial pressure}[/tex]
[tex]T_{1} = \text{initial temperature}[/tex]
[tex]P_{2} = \text{final pressure}[/tex]
[tex]T_{2} = \text{final temperature}[/tex]
Solution:
Starting with the formula of Gay-Lussac's law
[tex]\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}[/tex]
Multiplying both sides of the equation by T₁T₂ to eliminate the denominators
[tex]P_{1}T_{2} = P_{2}T_{1}[/tex]
Dividing both sides of the equation by P₁ to solve for T₂
[tex]T_{2} = T_{1} \times \dfrac{P_{2}}{P_{1}}[/tex]
Substituting the given values and solving for T₂
[tex]T_{2} = \text{290 K} \times \dfrac{\text{1900 mmHg}}{\text{19000 mmHg}}[/tex]
Therefore, the final temperature is
[tex]\boxed{T_{2} = \text{29 K}}[/tex]
Answer:
T₂ = 29 K
[tex]\\[/tex]
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