The speed of sound (v) in a gas might plausibly depend on the pressure (P), the density (p), and the volume V?

Question

(continuation)
of the gas. Use dimensional analysis to determine the exponents x, y, and z in the formula:
v= CP^xp^yV^z
where C is a dimensionless constant. Incidentally, the mks units of pressure are kilograms per meter per second squared (kg/ms^2).

Answer 1

The previous answer has an algebraic error where the x and y values are plugged into eq. 2:
-x -3y = 1, so z comes out = 0.
So the formula becomes
v = C * P^1/2 * p^-1/2 = C * sqrt(P/p)
These dimensions check out [sqrt(P/p) = sqrt((kg/(ms^2)) * (m^3/kg)) = sqrt(m^2/s^2) = m/s].
This is a relief, since the idea of velocity depending on volume did seem fishy. Watch out for textbook writers who refer to "plausible" ideas!

Answer 2

Dimensions of Lift Hand Side are :
M^0 L^1 T^-1

Dimensions of Right Hand Side :
[M^1 L^-1 T^-2]^x [M^1 L^-3 T^0]^y [M^0 L^3 T^0]^z
Open the brackets :
M^x L^-x T^-2x M^y L6-3y L^3z
Bring the like terms together and use the law of exponent :
M^(x + y) L^(3z - x - 3y) T^(-2x)

Dimensions of LHS = Dimensions of RHS

x + y = 0 .......................1
3z - x - 3y = 1 ......................2
-2x = -1 .......................3

Solve (3) :
x = 1/2
Put in (1) :
y = -1/2
Put in (2) :
z = 1/6

So the formula is :
v = C * P^1/2 * p^-1/2 * V^1/6 = C * sqrt. [(P * V^1/3) / p]

Hope this helps.

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