- #1
wongrufus
- 2
- 0
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
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
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