A sample of gas is at a pressure of 1 atm , a volume of 1 L , and a temperature of 25 degrees celsius. What would happen to the pressure if the volume were reduced to 0.5 L and the temperature increased to 250 degrees celsius?
Use
P1.V1/T1=P2.V2/T2
where T values are in K
To determine what happens to the pressure of a gas when the volume is reduced and the temperature is increased, we can use the combined gas law, which is derived from the ideal gas law.
The combined gas law is given by:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 and P2 are the initial and final pressures of the gas
V1 and V2 are the initial and final volumes of the gas
T1 and T2 are the initial and final temperatures of the gas
Now, let's plug in the values given in the question and solve for P2:
P1 = 1 atm
V1 = 1 L
T1 = 25 degrees Celsius = 298 K (converting to Kelvin scale)
V2 = 0.5 L
T2 = 250 degrees Celsius = 523 K (converting to Kelvin scale)
Using the formula, we can calculate P2:
(1 atm * 1 L) / 298 K = (P2 * 0.5 L) / 523 K
Let's solve for P2:
P2 = (1 atm * 0.5 L * 523 K) / (298 K * 1 L)
P2 = 0.8782 atm
Therefore, the pressure would be approximately 0.8782 atm when the volume is reduced to 0.5 L and the temperature is increased to 250 degrees Celsius.