I feel like the difference is likely to be negligible (unless omega is very very large), but here it goes anyway:

-Assuming the p is the pressure at the near face, not the far face or the average

-I am going to call d x instead. This reduces confusion when dealing with derivatives (because d could either be a variable or the sign for derivative).

Centripetal acceleration can be found from the rotation rate and distance from the center point:

a = omega^2 * (R + x)

This can give you the force:

F = m*a = m*omega^2 * (R+x)

The mass is the mass of a cylinder of air at distance d, with differential thickness dx.

m = pi * r^2 * dx * rho

F = pi * r^2 * dx * rho * omega^2 * (R+x)

Density is a function of pressure and temperature, which can be determined by the ideal gas law:

P = rho * Rspec * T

rho = P / (Rspec*T)*

F = pi * r^2 * dx * (P(x)/(RspecT)) * omega^2 * (R+x)

F is the net force on the differential cylinder, which depends only on the difference in pressure at distance x and distance x+dx. It would also depend on gravity if we accounted for it.

F = (pi * r^2) * (P (x + dx) - P (x)) = (pi * r^2) * ((P(x) + dP) - P(x))

F = (pi * r^2) * dP

Substitute in our expression for F:

F = pi * r^2 * dx * (P(x)/(Rspec*T)) * omega^2 * (R+x) = (pi * r^2) * dP

Divide by pi * r^2:

dx * (P(x)/(Rspec*T)) * omega^2 * (R+x) = dP

Define K = omega^2 / (Rspec*T) to clean up the left hand side:

dx * P(x) * K * (R+x) = dP

This is a linear differential equation for P(x), and can be solved by seperation:

dP/P(x) = dx * K * (R+x)

Integrate both sides:

ln(P(x)) = K*R*x + K*x^2 / 2 + C (Where C is constant of integration)

Exponential of both sides (define F = e^C):

P(x) = e^C * e^(K*R*x + K*x^2 /2) = F * e^(K*R*x + K*x^2 /2)

Use the boundary condition, P(0) = p to solve for f:

p = F * e^(0) * e^(0) = F

F = p

Substitute this back in for F

P(x) = p * e^(K*R*x + K*x^2 /2) = p * e^(omega^2 / (Rspec*T) * (R*x+x^2/2))

Using some actual numbers:

p = 120 psi

omega = 10 /s

T = 70 degrees Fahrenheit = 500 degrees Rankine

Rspec (for air) = 53.353 ft * lbf / (lb * degree Rankine)

R = 3 feet

x = 2 feet

P(0) = 120 psi

P(2) = 120.132 psi

So I don’t think you will get any significant pressure rise because of rotation, unless you are rotating extremely fast (10 rad/s is already more than a revolution per second). And rotating a tank very fast would seem to be a safety hazard. I’m curious, what is your application? An arm with the tank on it?