2d Flashcards
what can we do to change internal energy
we can change heat or work done
this is bc U = q + w
what is work
work is the energy expended by doing something
aka an expanding gas does work on the surroundings
think of a balloon pushing the surroundings away from it.
how do we find change in V ,, aka delta V,, aka change in volume
we do V2 - V1
is change in V positive or negative when finding change in v for an expanding gas
change in v will be positive
bc change in v = V2 - V1 and when things expand they get bigger, so V2 will be larger than V1,, meaning u will get a positive number.
how do we find change in work
- pext deltaV
pext = external pressure
delta V = change in volume
why is deltaW = -pext deltaV a negative value
bc the energy is going from the system to the surroundings,, bc ur doing the work onnnn the environment. ur the one doing the work so obvs the energy of u is gonna decrease.
when were on about change in work. aka deltaW,, are we thinking of the enrgy of the system or the energy of the environment
we are on about the energy of the system!!!
so if its doing work on the env,, its enrgy is decreasing bc ur reducing w.
and U = q+w
when we think of an ideal gas,, what do we assume
we assume that there are no intermolecular forces of attraction between the molecules
if there are no intramolecular forces of attraction between particles,, what else is missing
potential energy!! they have no potential energy,, only kinetic energy
if particles in an ideal gas have no potential energy,, meaning they only have kinetic energy,, what is their energy // kinetic energy proportional to
its proportional to temperature!!
aka an increase in temp will increase the particles kinetic energy
U.ideal is proportional to what
U.ideal is proportional to temp!!!
bc no intermolecular forces of attraction = only kinetic = increase temp = increase kinetic energy.
if we are told that the gas is ideal ,, and its in an isothermal condition,, what should we think
ideal gas = no intermolecular forcesof attraction so U.ideal is proportional to temp,, bc if no potential energy there is just kinetic energy.
and if there is no change in temp bc its isothermal ,, it means its kinetic energy doesnt change,, which means its internal energy also doesnt change.
so deltaV = 0
when we try to figure out ‘work done’ what must we think of
think that the system is giving energy to the surroundings. so its losing energy. so we think of the deltaW = -pext deltaV
okay what do we think when we see,, calculate work done + heate evolved if expansion is carried out in a vacuum
vacuum = pext = 0
so deltaW = -pext deltaV
so delta W is 0
delta q is also 0. bc deltaV = deltaQ + deltaW
energy of an ideal gas depends on what
tempppp
aka basically kinetic energy
no change in temp = no change in internal energy
find work done and heat evolved in expannsion if it occurs against a constant external pressure of 1atm.: what should we think
we need to find deltaV
finding volumes,, 1 and 2: using pV = NRT
do V2 - V1 to find deltaV.
then use delta V and corresponding pressure to find W. deltaW = -pext deltaV
deltaW = -pext NRT ( 1/p2 - 1/p1)
then put the value sin hereee to get deltaW
finding delta.q when u have deltaW and deltaU. delta = 0
delta q = (-1 x deltaW)
bc U = q + w
0 = q + w
q = 0 - w
so q = inverse of w
what should we think when we are asked to find the change in work and heat for a reversible reaction
the reaction will be in dynamic equilibrium with the surroundings
delta is used when
when there is a large change
sigma looking thing is used when
when there is a small change
aka in a reversible reaction where the system is in dynamic equilibrium with the surroundings.
when is d used
when there is an infinitesly small different. negligible difference.
delta w when we have volume
- pext delta V
- -NRT ln (v2 / v1)
in delta w = -nRT ln(v2 / v1) what is nrt
nrt is a constant meaning NRT can alxo we written assss
V= nRT/p
delta W when we have p
- NRT ln (p2/p1)
does a reversible reaction do more or less work than one at constant pressure
a reversible reaction does more work.
aka think of a straight line for a non reversible,, and a sloped line for reversible.
the sloped line has more shaded area underneath,, aka it does more work
