Answer:
The answer to your question is T2 = 610.9 °K
Explanation:
Data
Volume 1 = V1 = 30 cm³
Temperature 1 = T1 = 420 °K
Pressure 1 = P1 = 110 kPa
Volume 2 = V2 = 40 cm³
Temperature 2 = T2 = ?
Pressure 2 = P2 = 120 kPa
Process
To solve this problem use the combined gas law.
P1V1/T1 = P2V2/T2
-Solve for T2
T2 = P2V2T1 / P1V1
-Substitution
T2 = (120 x 40 x 420) / (110 x 30)
-Simplification
T2 = 2016000 / 3300
-Result
T2 = 610.9 °K
Answer:
477 °C
Explanation:
Step 1: Given data
- Initial pressure (P₁): 1.0 atm
- Initial temperature (T₁): 27 °C
- Final pressure (P₂): 2.5 atm
Step 2: Convert 27 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 27 + 273.15 = 300 K
Step 3: Calculate the final temperature (T₂)
If we assume constant volume (before the can explodes) and ideal behavior, we can calculate the final temperature using Gay-Lussac's law.
T₁/P₁ = T₂/P₂
T₂ = T₁ × P₂/P₁
T₂ = 300 K × 2.5 atm/1.0 atm = 750 K
In Celsius,
°C = K - 273.15 = 750 - 273.15 = 477 °C
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