a box containing 2.00m cubed of air is sealed at sea level pressure of 1.0 atm and a temperature of 20.0 degrees celsius. if the box is heated in an oven to a temperature of 250 degrees celsius, what is the final pressure in atm in the box?
assuming the air is ideal, we can use the formula,
P1/T1 = P2/T2
where
P1 = initial pressure
P2 = final pressure
T1 = initial temperature (in Kelvin)
T2 = final temperature (in Kelvin)
we first convert the given temp to Kelvin:
T1 = 20 + 273.15 = 293.15 K
T2 = 250 + 273.15 = 523.15 K
substituting,
P1/T1 = P2/T2
1/293.15 = (P2)/523.15
P2 = 523.15/293.15
P2 = 1.79 atm
hope this helps~ :)
To determine the final pressure in the box after heating, we can use the ideal gas law formula:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
First, we need to convert the temperature from Celsius to Kelvin. The formula for the conversion is:
T(K) = T(°C) + 273.15
Initial temperature, T1 = 20.0°C + 273.15 = 293.15 K
Final temperature, T2 = 250.0°C + 273.15 = 523.15 K
Now, we can calculate the initial and final number of moles of air in the box. We can use the ideal gas equation and the given volume of air.
Initial volume, V1 = 2.00 m^3
Final volume, V2 (which remains constant) = 2.00 m^3
Assuming that air behaves ideally, the number of moles in the box remains constant throughout the process.
n1 = n2
Finally, we can use the ideal gas law to solve for the final pressure, P2:
P1V1 = n1RT1
P2V2 = n2RT2
Since n1 = n2, we can simplify the equation to:
P1V1/T1 = P2V2/T2
Plugging in the values, we get:
(1.00 atm)(2.00 m^3)/(293.15 K) = P2(2.00 m^3)/(523.15 K)
P2 = (1.00 atm)(293.15 K)/(523.15 K) ≈ 0.564 atm
Therefore, the final pressure in the box after heating in the oven would be approximately 0.564 atm.