A gas that has a volume of 28.0 liters, a temperature of 45.0 °C, and an
unknown pressure has its volume increased to 34.0 liters and its
temperature decreased to 35.0 °C. If I measure the pressure after the
change to be 2.00 atm, what was the original pressure of the gas?
how would I solve this when I'm missing p1 and not one of the 2's?

3.12 atm

Explanation:

Data Given:

initial Volume of a gas V1= 28.0 L

Final Volume of gas V2= 34.0 L

initial Temperature of gas T1 = 45 °C

final Temperature of gas T2 = 35 °C

initial pressure of oxygen P1 = ?

final Pressure of oxygen P2= 2 atm

Solution:

we have to find the initial pressure of the gas as we are concerned with original pressure and for which it can be find by the following workup.

for this purpose

The combined Boyle's law and Charles's law formula will be applicable here at constant temperature

Combined gas law formula

                   P1V1 / T1 = P2V2 /T2

As we have to find initial pressure so rearrange the equation

                   P1 = (P2V2 / T2 ) x T1 / V1 (1)

Put values in Equation 1

                   P1 = (2.00 atm x 34 L / 35 °C) x 45 °C / 28 L

                   P1 = (1.94 atm L/°C) 45 °C / 28 L

                   P1 =  87.94 atm L / 28 L

                   P1 =  3.12 atm

So the original pressure was 3.12 atm

Note:

The given data is enough to calculate original pressure

As we know that according to Boyle's law pressure and volume has inverse relation as we increase the pressure the volume decreases and when we decrease the pressure the volume increases.

In the given scenario we have seen that volume increased so it means pressure should decrease also that is why the final pressure was 2 atm.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"