A sample of argon at 300.°C and 50.0 atm pressure is cooled in the same container to a temperature of 0.°C. What is the new pressure?
(P1V1/T1)=(P2V2/T2)
Don't forget to change T to Kelvin.
V1=V2 and isn't needed in the equation.
To find the new pressure of the argon gas after cooling, we can use the combined gas law, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law. The equation is as follows:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = Initial pressure (50.0 atm)
V1 = Initial volume (constant since the same container is used)
T1 = Initial temperature (300.°C + 273.15 = 573.15 K)
P2 = New pressure (unknown)
V2 = Initial volume (constant)
T2 = New temperature (0.°C + 273.15 = 273.15 K)
Rearranging the equation to solve for P2, we have:
P2 = (P1 * V1 * T2) / (V2 * T1)
Given that V1 = V2 is the same container, we can cancel out the volume terms:
P2 = (P1 * T2) / T1
Substituting the values into the equation:
P2 = (50.0 atm * 273.15 K) / 573.15 K
Calculating this expression:
P2 = 23.94 atm
Therefore, the new pressure of the argon gas after cooling to 0.°C is approximately 23.94 atm.
To find the new pressure, we can use the ideal gas law equation:
P1 * V1 / T1 = P2 * V2 / T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure (what we're trying to find)
V2 = final volume (we'll assume it stays constant)
T2 = final temperature
Given:
P1 = 50.0 atm
T1 = 300.°C = 300 + 273.15 = 573.15 K
T2 = 0.°C = 0 + 273.15 = 273.15 K
Since the volume remains the same, we can cancel it out:
P1 / T1 = P2 / T2
Now we can substitute the given values and solve for P2:
(50.0 atm) / (573.15 K) = P2 / (273.15 K)
Cross-multiplying:
P2 = (50.0 atm) * (273.15 K) / (573.15 K)
Calculating:
P2 ≈ 23.8 atm
Therefore, the new pressure of argon after cooling to 0.°C is approximately 23.8 atm.