Calculate the final pressure, in atmospheres, for each of the following with n and V constant.
A gas with an initial pressure of 1.20 atm at 82∘C is cooled to -28∘C.
The easy way to do this is to remember the gas laws. Pressure increases with increasing temperature. Therefore, if T is going from 82C to -28C you know it is being cooled which means the pressure will decrease. So make the factor so you get a smaller number. Remember T must be in kelvin.
Final P = 1.20 atm x (T factor)
To calculate the final pressure of the gas, we can use the combined gas law. The combined gas law relates the initial and final conditions of a gas when the number of moles (n) and the volume (V) are constant.
The combined gas law equation is given by:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 and P2 are the initial and final pressures, respectively,
V1 and V2 are the initial and final volumes, respectively,
T1 and T2 are the initial and final temperatures, respectively.
In this case, we are given the initial pressure (P1 = 1.20 atm), the initial temperature (T1 = 82°C), and the final temperature (T2 = -28°C). We need to find the final pressure (P2).
However, in the combined gas law, temperatures should be in Kelvin. So, we need to convert the temperatures from Celsius to Kelvin using the formula:
T (K) = T (°C) + 273.15
Converting the given temperatures, we have:
T1 = 82°C + 273.15 = 355.15 K
T2 = -28°C + 273.15 = 245.15 K
Now, we can substitute the known values into the combined gas law equation and solve for P2:
(1.20 atm * V1) / (355.15 K) = (P2 * V2) / (245.15 K)
Since the volume (V) is constant, we can simplify the equation:
1.20 atm / 355.15 K = P2 / 245.15 K
Now, we can cross-multiply and solve for P2:
1.20 atm * 245.15 K = 355.15 K * P2
294.18 atm*K = 355.15 K * P2
Dividing both sides by 355.15 K, we get:
P2 = (294.18 atm*K) / (355.15 K) ≈ 0.829 atm
Therefore, the final pressure of the gas is approximately 0.829 atm when cooled to -28°C.