I'm aware of the equation(adsbygoogle = window.adsbygoogle || []).push({});

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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# Change in pressure and Internal Energy

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