A constant volume of O2 gas is heated from 140C to 216C. The final pressure is 22.3 atm. What was the initial pressure?

P1/T1 = P2/T2

140 + 273 = T1
216 + 273 = T2

P1 = (T1/T2)22.3

To solve this problem, we can 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 in Kelvin.

First, we need to convert the temperatures from Celsius to Kelvin. Since the temperature difference is the same, it does not affect the final result.

Initial temperature (T1) = 140°C + 273.15 = 413.15K
Final temperature (T2) = 216°C + 273.15 = 489.15K

Since the volume is constant, we can cancel it out in the equation. The equation becomes:

(P1/T1) = (P2/T2),

where P1 is the initial pressure and P2 is the final pressure.

Now, let's substitute the given values into the equation:

P1/413.15K = 22.3 atm/489.15K,

To isolate P1, we can cross-multiply:

P1 = (22.3 atm x 413.15K) / 489.15K.

Calculating this expression gives us:

P1 = 18.84 atm.

Therefore, the initial pressure of the O2 gas was approximately 18.84 atm.