A sample of oxygen at 25 degrees C occupies 697 mL.
If this sample later occupies 1062 mL at 55 degrees C and 1.3 atm, what was its original pressure?
Answer in units of atm
(P1V1/T1) = (P2V2/T2)
Do I have to convert mL to L or just degrees to K?
To solve this question, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature measured in Kelvin.
First, let's convert the temperatures to Kelvin. To convert from Celsius to Kelvin, we add 273.15 to the Celsius temperature:
Temperature in Kelvin (Initial) = 25°C + 273.15 = 298.15 K
Temperature in Kelvin (Final) = 55°C + 273.15 = 328.15 K
Next, we use the given volumes and temperatures to find the ratio of initial and final pressures. According to Gay-Lussac's Law, at constant volume, the ratio of pressures is directly proportional to the ratio of temperatures:
(Pinitial / Tinitial) = (Pfinal / Tfinal)
We can rearrange this equation to solve for Pinitial:
Pinitial = (Pfinal * Tinitial) / Tfinal
Now, let's substitute the values into the equation:
Pinitial = (1.3 atm * 298.15 K) / 328.15 K
Calculating this, we find:
Pinitial ≈ 1.178 atm
Therefore, the original pressure of the sample of oxygen was approximately 1.178 atm.