To solve this problem, we can use the combined gas law equation:
P1V1/T1 = P2V2/T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Given:
P1 = 92.0 kPa
V1 = 500 mL = 0.5 L
T1 = 28.0oC = 28.0 + 273.15 = 301.15 K
V2 = 300 mL = 0.3 L
T2 = -5.0oC = -5.0 + 273.15 = 268.15 K
Now we can plug these values in and solve for P2:
P1V1/T1 = P2V2/T2
(92.0 kPa)(0.5 L)/(301.15 K) = P2(0.3 L)/(268.15 K)
46.0/301.15 = P2/268.15
0.1527 = P2/268.15
P2 = 0.1527 * 268.15
P2 = 40.98 kPa
Therefore, the pressure exerted by the sample when compressed to 300 mL and cooled to -5.0oC will be approximately 40.98 kPa.
In an investigation of the properties of the coolant gas used in an air-conditioning system, a
sample of volume 500 mL at 28.0oC was found to exert a pressure of 92.0 kPa. What pressure
will the sample exert when it is compressed to 300 mL and cooled to -5.0oC? Leave your answer
in kPa.
1 answer