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I need to check units in this equation, I know, that there's Pa in the numerator, but what about the denominator?

$p_1$ - pressure in [Pa]
$T_1$, $T_2$ - temperature in [K]
$\rho_w$ - density
$g$ - acceleration

Units

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  • $\begingroup$ thee answer is metre because density is in kg/m^3 and acceleration is in m/s^2. $\endgroup$ – Abhimanyu Pallavi Sudhir Jun 6 '13 at 13:34
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HINT:

You know that $\rho gh$ is defined as pressure inside a liquid at a depth $h$ . Now you can get dimensions of denominator. $$\rho gh\equiv[Pa]$$

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Density is mass per volume, so that we get: $\dfrac{kg}{m^3}$

Acceleration is $\dfrac{m}{s^2}$

The two multiplied together gives you: $\dfrac{kg}{s^2m^2}$

Now, if you want to turn this into the SI units of pressure, you can, since the units of pressure can be derived from depth, density and acceleration due to gravity ($P = h\rho g$) and you get: $$\text{Units of pressure}: \text{Pa} = m\times \dfrac{kg}{m^3} \times \dfrac{m}{s^2}$$

You can recognize the second part, so it becomes $\dfrac{\text{Pa}}{m} = \dfrac{kg}{s^2m^2}$

Above: $\text{Pa}$;

Below: $\dfrac{\text{Pa}}{m}$

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    $\begingroup$ Since when is density volume per mass?? As far as I know, it's always been the other way around. If you express pascal using force per area, you get $\mathrm{kg}\cdot\mathrm{m}^{-1}\cdot\mathrm{s}^{-2}$. $\endgroup$ – Ondřej Černotík May 26 '13 at 14:11
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    $\begingroup$ @OndřejČernotík Indeed! The final answer (above/below) is correct again though because the mistake cancels out. $\endgroup$ – Michiel May 26 '13 at 14:45
  • $\begingroup$ Sorry, it was a brain fart >~< Thanks for pointing it out! $\endgroup$ – Jerry May 26 '13 at 17:18

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