Question

The derivation of the equation of adiabatics process: deltaQ=deltaU+W, deltaQ=0, deltaU+W=0, deltaU=-W, nC(v,m)(deltaT)=-p(delta V). Problem: why the work done=p(delta V) since the pressure of the gas is not constant?

Answer 1

while deriving this expression we considered a piston compressed to create a uniform force so pressure was constaant
as pressure=force/area
area being the same

Answer 2

but if the pressure is not constant?

Answer 3

why shouldnt it be constant?...it is only under constant pressure that the relation of thermodynamic law 1 is derived

remember when pressure is different then at say same volume then
qv=delta u

Answer 4

But how to look at the pressure... For example, if in a adiabatic process. The Pressure is not constant right? But it uses its internal energy to do the workdone.... so, if we just calculate the work done... pdV... we assume the p is constant... wont that be contradicted?

Answer 5

while using 1 thermodynamic equation, we have assumed pressure to constant
adiabation means no energy enters or leaves the system meaning delta q=0
so w=delta u
which means with change in internal energy u
the delta v (volume)changes
p*deltav=delta u

Answer 6

trying to understand

Answer 7

But in general, you're right. Pressure is a function of volume, P = P(V). That is why in the general case, we define the work not to be just
\[ P \Delta V \] but as the sum over the change in volume:
\[ Work = \int_{P_A}^{P_B} P \ dV \]

Answer 8

*limits should be V_A to V_B

Answer 9

OK. confused

Answer 10

here also we assume pressure as constant as it is defined already while proving this concept

Answer 11

hmm
, thanks salini.