By how much does the air pressure in a house increase if the house is sealed and the air temperature increases from 10.7° C to 21.0° C? The initial air pressure is 1.00 atmosphere.
To calculate the change in air pressure in a sealed house, we can use Charles's Law. Charles's Law states that for a given amount of gas at a constant pressure, the volume of the gas is directly proportional to its temperature.
The formula for Charles's Law is:
(V1 / T1) = (V2 / T2)
Where:
V1 = initial volume
T1 = initial temperature
V2 = final volume
T2 = final temperature
Since the house is sealed, the volume remains constant. So, we can rewrite the formula as:
T1 / T2 = P2 / P1
Where:
P1 = initial pressure
P2 = final pressure
Given:
T1 = 10.7°C = 283.85 K (Convert to Kelvin)
T2 = 21.0°C = 294.15 K (Convert to Kelvin)
P1 = 1.00 atmosphere
Now, we can plug in the values into the formula:
T1 / T2 = P2 / P1
283.85 K / 294.15 K = P2 / 1.00 atmosphere
Cross-multiplying:
P2 = 1.00 atmosphere × (294.15 K / 283.85 K)
P2 = 1.0362 atmospheres
Therefore, if the air temperature in a sealed house increases from 10.7°C to 21.0°C, the air pressure will increase by approximately 0.0362 atmospheres.