How would I go about answering the challenge problem? I got 549 torr as the answ

Question

How would I go about answering the challenge problem?

I got 549 torr as the answer for question 2

Chemistry 152 1. Convert: (a) 357.0 torr to atm; (b) 5.72 kPa to torr: (c) 0.30 Pa to atm 2. An open-ended mercury-filled manometer is co Homework # 1.1 pen-ended mercury-filled manometer is connected to a bulb of Ar. If the mercury on the open side is found to be mospheric pressure is 0.915 atm what s the pressure of the AI 14.6 cm lower than the mercury on the Ar side, and the at pressure of Ar be in the problem 2 if the manometer was filled with a hydrocarbon with Challenge Problem: What would the density 1.458 g/mL instead of mercury? (i.e. use the same problem description, same instead of mercury? (i.e. use the same problem description, same measurements)

Explanation / Answer

Q2

Argon;

hh = 14.6 cm lower

P atm = 0.915 atm

find P ar

Patm = Pcolumn + Pgas

Pcolum = rho*g*h = 13600 * 9.8 * 14.6/100 = 19458.88 Pa = 19458.88/101325 atm = 0.192044 atm

Pgas = Patm - Pcolumn = 0.915-0.192044 = 0.722956 atm

change atm to torr

1 atm = 760 torr

Pgas = 0.722956 atm --> 0.722956*760 torr = 549.44656 torr, as you stated

Challenge problem:

Find P of Argon given it is filled with Hydrocarbon, insted of mercury

so, simply, assume the SAME height

only density changes

Patm = Pcolumn + Pgas

D = 1.458 g/mL = 1458 kg/m3

Pcolum = rho*g*h = 1458* 9.8 * 14.6/100 = 2086.1064 Pa = 2086.1064 /101325 atm = 0.02058atm

Pgas = Patm - Pcolumn = 0.915-0.02058 = 0.89442 atm

change to torr

0.89442 --> 0.89442*760

Pgas = 679.7592 torr

it must decrease, since the density of fluid is much less

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