A 32.3-L of a gas at 30 degrees Celsius and 1.20 atm is to be cooled at constant pressure until its volume becomes 28.4 L. What will be the new gas temperature?
(V1/T1) = (V2/T2)
Remember T must be in kelvin
To find the new gas temperature, we can use the combined gas law, which states that the ratio of the initial pressure, volume, and temperature to the final pressure, volume, and temperature is constant.
The combined gas law is written as:
(P1 * V1) / T1 = (P2 * V2) / T2
Given:
Initial volume (V1) = 32.3 L
Initial temperature (T1) = 30 degrees Celsius = 30 + 273.15 = 303.15 K
Initial pressure (P1) = 1.20 atm
Final volume (V2) = 28.4 L
Final pressure (P2) = same as initial pressure, since the problem specifies constant pressure
Now, let's plug in the known values into the combined gas law equation and solve for the final temperature (T2):
(1.20 atm * 32.3 L) / 303.15 K = (1.20 atm * 28.4 L) / T2
Simplifying the equation:
(38.76 atm·L) / 303.15 K = (34.08 atm·L) / T2
Cross-multiplying and solving for T2:
38.76 atm·L * T2 = 34.08 atm·L * 303.15 K
T2 = (34.08 atm·L * 303.15 K) / 38.76 atm·L
T2 ≈ 267.43 K
Therefore, the new gas temperature will be approximately 267.43 Kelvin.