5.1b PV Work

Question 1

PV Work (pressure volume work)

Work will not come out in J if you use the standard equation shown below
* w = -Pext * Delta V

Question 2

Enthalpy = the total heat content of a system at a constant pressure

It includes:
* The system’s internal energy

Delta H = heat absorbed or released at cosntant pressure

Enthalpy is a way to track heat flow in/out of a chemical system when pressure is constant (its the change in internal energy)
* So when a chemical reaction happens, Delta H tells you whether the system released heat (exothermic) or sucked it up (endothermic)

Question 3

Work = Force * Displacement

Work = -Pext * Delta V
* Thats P at constant external pressure * the change in volume

Did not fully understand the below, but lots of important equations are derivations

Answer 3

Question 4

Calculate the change in internal energy if 100J of heat is transferred from the surroundings to the sample of gas (the system) and the gas expands, increasing in volume by 2L under a constant external pressure of 1ATM

So heat went into the system (100J of heat)

then were told this gas, at a constant ATM expanded by 2L
* since its an expansion we know that work is going to be negative, however we still need to figure out the value of work

We can’t just plug these numbers into the equation for work below
* w = -Pext * delta volume
* we can’t just plug it into here because it doesnt come out in jouls
* can’t just do -(1 atm)(2L) - doesnt come out in jouls

1J = Si unit for energy
SI unit for pressure is the pascle (not ATM)
SI unit for volume = m^3

However, if we convert our ATM to pascles and our L into m^3, then we can plug them into that equation and it will come out in jouls

**1 ATM = 101,325 Pascles

** 1000L = 1m^3 **

w = - (101,325Pa * 0.002m^3) = -202.65 J

Now we can plug into our delta E equation
* Delta E = q + w
* Delta E = 100J - 202.65J
* Delta E = -102.65J

Second route:

Universal gas constant
* can be provided 2 different ways
* R = 0.08206 L * atm / mol * k <— common
* R = 8.314 J / mol * kelvin

0.08206 L * atm = 8.314 J
* we could use this as a conversion factor
* shown below
* I dont love this method, Im going to use the first one

Question 5

Answer 5

-30J

Question 6

Answer 6

-190 J

Remember, system doing work on surroundings = negative work
* whenever something is doing work on something else its losing the work and the something else (surroundings in this case) is gaining the work
* always from the perspective of the system

Question 7

Answer 7