To solve this problem, we can use the combined gas law formula:
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 degrees Celsius = 301 K
V2 = 300 ml = 0.3 L
T2 = -5.0 degrees Celsius = 268 K
Now plug the values into the formula:
(92.0 kPa)(0.5 L)/(301 K) = P2(0.3 L)/(268 K)
P2 = (92.0 kPa)(0.5 L)(268 K)/(301 K)(0.3 L)
P2 = 51.6 kPa
Therefore, the pressure of the cobalt gas sample will be 51.6 kPa when compressed to 300 ml and cooled to -5.0 degrees Celsius.
In an investigation of the properties of the cobalt gas used in an air conditioning system , a sample of volume 500ml at 28 degrees Celsius was found to exert a pressure of 92.0 kpa . What pressure will be the sample exert when it is compress to 300 ml and cool to -5.0 degrees Celsius? Leave your answer in kpa.
1 answer