Question

Problem 10.19 The typical atmospheric pressure on top of Mt. Everest (29,028 ft) is about 265 torr. Convert this pressure to

Part A atm 760   atm   SubmitMy AnswersGive Up Incorrect; Try Again Part B mmHg 1   mmHg   SubmitMy AnswersGive Up Incorrect; Try Again Part C pascals 3.53×104   Pa   SubmitMy AnswersGive Up Correct Significant Figures Feedback: Your answer 3.52⋅104 = 3.52×104 Pa was either rounded differently or used a different number of significant figures than required for this part. Part D bars 0.353   bar   SubmitMy AnswersGive Up Correct Significant Figures Feedback: Your answer 0.352 bar was either rounded differently or used a different number of significant figures than required for this part. Part E psi   psi   SubmitMy AnswersGive Up

Answer 1

1 atm = 760 torr = 14.7 psi.= 1.013 bar = 101.325 kPa= 760mm of Hg

PART A:

265 torr = ? atm

We know that    760torr = 1 atm

then                                    265 torr = ? atm

                            = 265torr *1atm/760 torr = 0.349 atm

PART B:

265 torr = ? mm of Hg

We know that 760torr = 760mm of Hg

   then    265 torr = ? mm of Hg

=        ( 265 torr * 760mm of Hg )/ 760torr                  

= 265 mm of Hg

PART C:

265 torr = ? pascals

We know that     760 torr---------------- 1.013 *10^5 pa

                          265 torr -----------------? pa

       = (265torr * 1.013 *10^5 pa)/760 torr

       = 3.53 * 10^4 pascal

PART D:

265 torr = ? bar

We know that     760 torr---------------- 1.013 bar

     then               265 torr------------------ ? bar

                      = (265 torr * 1.013 bar)/760 torr

                      = 0.353 bar

PART E:

265 torr = ? psi

We know that          760 torr = 14.7 psi

                                265 torr = ? psi

                            = (265 torr* 14.7 psi)/760 torr

                            = 5.13 psi