# Change in pressure and Internal Energy

1. Mar 5, 2012

### wongrufus

I'm aware of the equation
ΔU = Q + W
ΔU is the change in internal energy which equals 3/2 nRΔT
Q is heat transfer
W is workdone

so PV = nRT
would it be correct to say that
ΔP V = nRΔT?

For isovolumetric process (i.e. no volume change, no work done),
can i say that
change in internal energy = 3/2 nRΔT = 3/2 ΔP V?

Say
Pressure changes from 200Pa --> 100Pa
Volume remains at 6 m^3
Would it be right to conclude that the change in internal energy = 3/2 x 100 x 6 = 900J?

**New Question: Is it possible (and if possible, under what condition) that the initial PV and final PV are known BUT the change in internal energy cannot be found directly by ΔU = 3/2 ΔPV?
coz i always come across questions where a P-V graph is given but the internal energy has to be calculated by more complicated methods,
i.e. ΔU = Q + W and not ΔU = 3/2 ΔPV

thx*c

Last edited: Mar 6, 2012
2. Mar 5, 2012

### apb000

I believe that is correct for a monatomic ideal gas. 900 J of heat flows into the system, which raises the temperature, which increases the pressure, which accounts for the increase in internal energy by the relation ΔU=3/2 ΔP V as you say.

3. Mar 6, 2012

### wongrufus

does the equation
ΔU = 3/2 nRΔT = 3/2 PΔV = 3/2 ΔPV
have its limitations?