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Q26P

Expert-verifiedFound in: Page 408

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**To suck lemonade of density ${\mathbf{1000}}{\mathbf{kg}}{\mathbf{/}}{{\mathbf{m}}}^{{\mathbf{3}}}$ up a straw to a maximum height of ${\mathbf{4}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{cm}}$, what minimum gauge pressure (in atmospheres) must you produce in your lungs?**

Gauge pressure produced in the lungs in atmospheric unit is $-3.9\times {10}^{3}\mathrm{atm}$

Maximum height, $\mathrm{h}=4\mathrm{cm}\text{or}0.04\mathrm{m}$

Density of lemonade, $\rho =1000\mathrm{kg}/{\mathrm{m}}^{3}$

Acceleration due to gravity, $g=9.8\mathrm{m}/{\mathrm{s}}^{2}$

**To find gauge pressure in the atm, we can use the formula for pressure in terms of density, acceleration due to gravity, and height. Using the conversion factor, we can convert the pressure into atmospheric pressure.**

Formula:

Pressure applied on a body, $\mathrm{p}=\mathrm{\rho gh}$

The pressure produced in the lungs in SI unit using equation (i) can be given as follows:

$p=-1000\mathrm{kg}/{\mathrm{m}}^{3}\times 0.04\mathrm{m}\times 9.8\mathrm{m}/{\mathrm{s}}^{2}$

$=-392\mathrm{Pa}$

role="math" localid="1657535402885" $=-392\mathrm{Pa}\times \frac{1\mathrm{atm}}{1.01\times {10}^{5}\mathrm{Pa}}(1\mathrm{atm}=1.01\times {10}^{5}\mathrm{Pa})$

$=-3.9\times {10}^{-3}\mathrm{atm}$

So, the pressure value is $-3.9\times {10}^{-3}\mathrm{atm}$

Negative sign indicates that the pressure inside the lung is less than the outside pressure.

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