Chapter 11, Problem 11.23QP

Interpretation Introduction

Interpretation:

The given mmHg conversion into atmospheres, bars, torr, and pascals has to be explained.

Concept Introduction:

• $\text{760}\text{.0 mmHg}$ equals $\text{101}\text{.325 kPa,}$ as a result both values will be involved. This location is slightly curious since most conversions involve a one, usually in the denominator.  The conversion examples above are examples of a one being involved.  For example, $\text{760}\text{.0 mmHg / 1}\text{.00 atm in examples 1 and 2}\text{. The 1}\text{.00}$ was assumed to be present.
• One atm equals $\text{760}\text{.0 mmHg}$, thus there will be a multiplication or division based on the direction of the change.
• One atm equals $\text{101}\text{.325 kPa,}$ accordingly there will be a multiplication or division based on the direction of the change.

Answer to Problem 11.23QP

The given pressure in $\text{atm = 0}\text{.764}$

The given pressure in $\text{kPa = 77}\text{.5}\text{kPa}$

The given pressure in $\text{bar = 0}\text{.775}\text{bar}$

Explanation of Solution

Record the given data

Given pressure is $\text{375 mmHg}$

$\text{1}\text{atm}\text{=}\text{101}\text{.325}\text{Pa}$

Given pressure $\text{375 mmHg}$is converting into atm

$\text{?atm}\text{=}\text{375}\text{mmHg}\text{×}\frac{\text{133}\text{.322}\text{Pa}}{\text{1}\text{mmHg}}\text{×}\frac{\text{1}\text{atm}}{\text{101}\text{.325}\text{pa}}\text{=}\text{0}\text{.493}\text{atm}$

$\text{?atm}\text{=}\text{375}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\text{=}\text{0}\text{.493}\text{atm}$

The pressure in atm is calculated by plugging in the values of the given pressure and $\text{1}\text{atm}\text{=}\text{101}\text{.325}\text{Pa}$.  The pressure in atm is found to be $\text{0}\text{.493}\text{atm}$

Given pressure $\text{375 mmHg}$is converting into bar

$\text{?bar}\text{=}\text{375}\text{mmHg}\text{×}\frac{\text{133}\text{.322}\text{Pa}}{\text{1}\text{mmHg}}\text{×}\frac{\text{1}\text{bar}}{\text{1×}{\text{10}}^{\text{5}}\text{Pa}}\text{=}\text{0}\text{.500}\text{bar}$

The pressure in bar is calculated by plugging in the values of the given pressure $\text{375 mmHg}$.  The pressure in bar is found to be $\text{0}\text{.500}\text{bar}$

Given pressure $\text{375 mmHg}$is converting into torr

$\text{?torr}\text{=}\text{375}\text{mmHg}\text{×}\frac{\text{133}\text{.322}\text{Pa}}{\text{1}\text{mmHg}}\text{×}\frac{\text{1}\text{torr}}{\text{133}\text{.322 Pa}}\text{=}\text{375torr}$

Note that because both $\text{1 mmHg and 1 torr are equal to 133}\text{.322 Pa,}$ this could be simplified by recognizing that $\text{1 torr = 1 mmHg}\text{.}$

$\text{375}\text{mmHg}\text{×}\frac{\text{1}\text{torr}}{\text{1 mmHg}}\text{=}\text{375}\text{torr}$

The pressure in $\text{torr}$ is calculated by plugging in the values of the given pressure is $\text{375 mmHg}$.  The pressure in torr is found to be $\text{375}\text{torr}$

Given pressure $\text{375 mmHg}$is converting into bar

$\text{?Pa}\text{=}\text{375}\text{mmHg}\text{×}\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\text{×}\frac{\text{101}\text{.325}\text{Pa}}{\text{1}\text{atm}}\text{=}\text{5}\text{×}{\text{10}}^{\text{4}}\text{Pa}$

The pressure in Pa is calculated by plugging in the values of the given pressure $\text{375 mmHg}$.  The pressure in Pa is found to be $\text{5}\text{×}{\text{10}}^{\text{4}}\text{Pa}$

Conclusion

The given mmHg is converted into atmospheres, bars, torr, and pascals was explained.