Question

An isochoric process is a process that happens at a constant Volume (i.e. in a rigid tank). However, the density of the fluid changes with pressure. Why do we draw on a P-v (Pressure - Specific Volume) a constant process, that is, constant specific volume? Can anyone give me a good explaninantion using relation formulas? Thank you!

Answer 1

Start with the first law of thermodynamics:
dQ = dU + dW

(remember that the d's on Q and W are "inexact differentials", if you really care at all)

By definition, there is no heat transfer in an adiabatic process. Thus, any infinitesimal heat transfer, dQ, is zero. Thus:

0 = dU + dW

dW we can express in terms of pressure and volume:
dW = P*dV

Thus:
0 = dU + P*dV

Expressing U, the internal energy, in terms of temperature. We assume that the gas is CALORICALLY PERFECT. That is to say, that (constant volume basis) specific heat capacity, cv, isn't dependent upon temperature.

U = n*cv*T

where n is the number of moles of gas
cv is the molar specific heat, on a constant volume basis
T is the temperature in Kelvin

It doesn't really matter where we offset the zero point of internal energy, so by default we set it equal to zero when T = absolute zero (if such a state really did exist for gas phase).

Use the ideal gas law, and solve it for n*T:
P*V = n*R*T

n*T = P*V/R

Thus:
U = cv*P*V/R

Take derivative relative to volume. Remember to use the product rule, since pressure is a function of volume. R and cv are constants, so they can be pulled out in front.
U = (cv/R) * P*V

dU/dV = (cv/R) * (P*dV/dV + V*dP/dV)
dU/dV = (cv/R) * (P + V*dP/dV)

Multiply through by dV:
dU = (cv/R) * (P + V*dP/dV) * dV
dU = (cv/R) * (P*dV + V*dP)

Substitutue back in:
0 = (cv/R) * ( P*dV + V*dP) + P*dV

Distribute:
0 = P*dV*(cv/R + 1) + V*dP*(cv/R)

Split the equation:
P*dV*(cv/R + 1) = -V*dP*(cv/R)

Gather the coefficients:
P*dV*(-(cv/R + 1)/(cv/R)) = V*dP
P*dV*(-(1 + R/cv)) = V*dP

----------------------
Now to consolidate our coefficient terms, containing cv in them.

Recall how specific heats of a gas are defined:

cp is defined from enthalpy:
H = n*cp*T

Which is by definition:
H = U + P*V

cv is defined from internal energy:
U = n*cv*T

Thus:
n*cp*T = n*cv*T + P*V

Replace P*V with the ideal gas law term:
n*cp*T = n*cv*T + n*R*T

Cancel n and T:
cp = cv + R

And the ratio gamma, is defined as:
gamma = cp/cv

Solve two equations, two unknowns, for cv and cp:
cp = gamma*R/(gamma + 1)
cv = R/(gamma + 1)

We are trying to simplify the term, (-1 - R/cv). Substitute cv accordingly.

(-1 - R/(R/(gamma+ 1)))
(-1 - (gamma + 1))
-gamma

Thus:
-1 - R/cv = -gamma
---------------------------------------...

Substitute accordingly:
-gamma*P*dV = V*dP

Gather like variables with like differentials:
-gamma*dV/V = dP/P

Recall how to integrate dx/x to ln(x)

-gamma*ln(V) = ln(P) + constant

Exponentiate both sides:
exp(-gamma*ln(V)) = exp(ln(P) + constant)

Simplify:
-gamma*ln(V) = ln(V^(-gamma))
exp(ln(V^(-gamma))) = V^(-gamma)

exp(ln(P) + constant) = exp(constant) * exp(ln(P)) = exp(constant) * P

exp(constant) is still a constant. Redefine accordingly.
exp(ln(P) + constant) = constant*P

Thus:
V^(-gamma) = P*constant

Recall:
V^(-gamma) = 1/V^gamma

1/V^gamma = P*constant

1 = constant*P*V^gamma

Move the constant to the other side, redefine, and get our result:
P*V^gamma = constant

There you have it.
P*V^gamma = constant for isochronic process