What will be the new pressure when a volume of 21.0 L of a gas at 35.0 degrees celsius and 2.00 atm is compressed to 12.0 L and cooled to 10.0 degrees celcius with no change in the number of moles?
Use (P1V1/T1) = (P2V2/T2)
Don't forget T is in Kelvin.
To solve this problem, we can use the combined gas law, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law. The equation is given as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
P2 = final pressure
V1 = initial volume
V2 = final volume
T1 = initial temperature (in Kelvin)
T2 = final temperature (in Kelvin)
Now let's solve the problem step by step using the given values:
Step 1: Convert the temperatures to Kelvin.
The temperature given is in Celsius, and to use the gas laws, we need to convert it to Kelvin. The formula to convert Celsius to Kelvin is: T(K) = T(°C) + 273.15.
T1 = 35.0°C + 273.15 = 308.15 K
T2 = 10.0°C + 273.15 = 283.15 K
Step 2: Plug the values into the combined gas law equation.
(P1 * V1) / T1 = (P2 * V2) / T2
(2.00 atm * 21.0 L) / (308.15 K) = (P2 * 12.0 L) / (283.15 K)
Step 3: Solve for P2.
Cross multiply and solve for P2:
(2.00 atm * 21.0 L * 283.15 K) = (P2 * 12.0 L * 308.15 K)
P2 = (2.00 atm * 21.0 L * 283.15 K) / (12.0 L * 308.15 K)
P2 ≈ 1.54 atm
So, when the volume is compressed to 12.0 L and cooled to 10.0 degrees Celsius, the new pressure will be approximately 1.54 atm.