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The pressure of the gas when it's temperature reaches 928 °C is 3823,36 kPa
To solve that we need to apply Gay-Lussac's Law. It states that the pressure of a gas when the volume is left constant (like in the case of a sealed container like an aerosol can) is proportional to temperature. This is the relationship derived from this law that we use to solve this problem:
[tex]P2= \frac{P1}{T1}*T2= \frac{103 kPa}{25}*928=3823,36 kPa[/tex]
Have a nice day!
The pressure of the gas when it's temperature reaches 928 °C is 3823,36 kPa
To solve that we need to apply Gay-Lussac's Law. It states that the pressure of a gas when the volume is left constant (like in the case of a sealed container like an aerosol can) is proportional to temperature. This is the relationship derived from this law that we use to solve this problem:
[tex]P2= \frac{P1}{T1}*T2= \frac{103 kPa}{25}*928=3823,36 kPa[/tex]
Have a nice day!
Answer:
415.11 kPa
Explanation:
The gas law to be applied here is Gay-Lussac's law. Gay-Lussac's law states that the pressure of a given mass of gas is directly proportional to it's temperature (in kelvin) provided volume remains constant.
The formula for the law is
P₁/T₁ = P₂/T₂
where P₁ is initial pressure (103 kPa)
P₂ is final pressure (unknown)
T₁ is initial temperature (25°C + 273 = 298K)
T₂ is final temperature (928°C + 273 = 1201K)
From the formula,
103/298 = P₂/1201
P₂ = [tex]\frac{103 * 1201}{298}[/tex]
P₂ = 415.11 kPa
415.11 kPa is the pressure of the gas when it temperature reaches 928°C.