# Question eead2

Apr 22, 2015

The greatest pressure is $\text{C)}$ $\text{12.0 psi}$.

I am going to convert all of the pressures into Pascals (Pa).

A. 0.713 atm

$\text{1 atm=101325 Pa}$

0.713 cancel("atm") xx (101325 "Pa")/(cancel(1 "atm"))=72200 "Pa"

B. 345 mmHg

$\text{1 mmHg=133.3224 Pa}$

345 cancel("mmHg")xx(133.3224 Pa)/(1 cancel("mmHg"))=46000 "Pa"

C. 12.0 psi

$\text{1 psi=6894.757 Pa}$

12.0cancel("psi")xx(6894.757 "Pa")/(1cancel("psi"))=82700 "Pa"

D. 58468 Pa (Needs no conversion)

E. 16.19 inHg

$\text{1 inHg=3386.389 Pa}$

16.19cancel("inHg")xx(3386.389 "Pa")/(1cancel("inHg"))=54830 "Pa"

The greatest pressure is C. 12.0 psi . When converted to Pascals, it is 82700 Pa, which is the greatest number of Pascals for all five values given in the problem.

To get the conversion factors I used the NIST website: http://physics.nist.gov/Pubs/SP811/appenB8.html

All of the calculated conversions were rounded to the correct number of significant figures.

Apr 22, 2015

The answer is C. You will need to find a table of conversion factors (or online app) answer this question. The conversion factors are written as fractions and multiplied to obtain the desired unit. Note that the original unit is in the denominator each time.

Convert all to a common unit to compare such as atmospheres (atm)

A 0.713 atm (no conversion required)

B. $345 m m H g \cdot \frac{1 a t m}{760 m m H g} = 0.454 a t m$

C. 12.0 "psi" * (1atm)/(14.7 "psi") = 0.816 atm 

D. $58486 P a \cdot \frac{1 a t m}{101325 P a} = 0.577 a t m$

E. 16.9 "in Hg" * (1atm)/(29.9 " in Hg") = 0.565 atm #