Question

1. A student reads a barometer in the laboratory and finds the prevailing atmospheric pressure to be 699 mm Hg. Express this pressure in atmospheres, kilopascals, torrs, pounds per square inch, and pascals.

Hint: 1 atm = 101.3 kPa = 760 torr = 760 mm Hg = 14.69 psi = 1.013×10⁵ Pa

mm Hg	atm	kPa	torr	psi	Pa
699

2. An automobile tire is inflated to a pressure of 29.2 psi. Express this pressure in atmospheres, kilopascals, inches Hg, millimeters Hg and torr.

Hint: 1 atm = 101.3 kPa = 29.92 in Hg = 760 mm Hg = 14.69 psi = 760 torr

psi	atm	kPa	in Hg	mm Hg	torr
29.2

3. Liquid isopropyl iodide has a density of 1.70 g/mL. A barometer is constructed using isopropyl iodide instead of mercury. If the atmospheric pressure is 0.964 atm, what is the height of the isopropyl iodide column in the barometer in cm?

The density of liquid mercury is 13.6 g/mL.

______ cm

Answer 1

Answer: 1 In the laboratory, a student calculates the prevailing atmospheric pressure using a barometer. The prevailing atmospheric pressure calculated by the student, P= 699 mm Hg. The relation between the different unit conversions of pressure is as follows: 1 atm =101.3 kPa = 760 torr = 760 mm Hg = 14.69 psi =1.013x10' Pa Calculate the prevailing atmospheric pressure (P in atm) as shown below: P=699 mm Hg The unit conversion is as follows: 1 atm = 760 mm Hg 1 atm P=699 mmHg x=0 8 * 760 mm Hg P=0.919 atm

Therefore, the prevailing atmospheric pressure, P is 0.919 atm). Calculate the prevailing atmospheric pressure (P in kPa) as shown below: P=699 mm Hg The unit conversion is as follows: 101.3 kPa = 760 mm Hg P=699 mmHex101.3 kPa 113 * 760 mmHg P=93.169 kPa P-93.2 kPa Therefore, the prevailing atmospheric pressure, P is 93.2 kPa. Calculate the prevailing atmospheric pressure (P in torr) as shown below: P=699 mm Hg

The unit conversion is as follows: 760 torr = 760 mm Hg . 760 torr P=699 mm Hg x= ** 760 mmHg P=699 torr Therefore, the prevailing atmospheric pressure, P is 699 torr). Calculate the prevailing atmospheric pressure (P in psi) as shown below: P=699 mm Hg The unit conversion is as follows: 760 mm Hg = 14.69 psi 14.69 psi P=699 mmHg x ng 760 mmHg P=13.510 psi P = 13.5 psi

Therefore, the prevailing atmospheric pressure, P is 13.5 psi. Calculate the prevailing atmospheric pressure (P in Pa) as shown below: P=699 mm Hg The unit conversion is as follows: 760 mm Hg =1.013x10' Pa P=699 mmHg 1.013x10* Pa 13 * 760 mmHg P=93169 Pa P=93.169x103 Pa P=93.2x10 Pa Therefore, the prevailing atmospheric pressure, P is 93.2x10² Pal.

Therefore, the prevailing atmospheric pressure of 699 atm in different units of pressure is as follows: mm Hg 699 atm 0.919 kPa torr | psi Pa 93.2 699 13.5 | 93.2 x103 Answer:2 An automobile tire is inflated to a pressure (P) of 29.2 psi. The relation between the different unit conversions of pressure is as follows: 1 atm = 101.3 kPa = 29.92 in Hg = 760 mm Hg = 14.69 psi = 760 torr ...... (2) Calculate the inflated pressure (P in atm) as shown below: P= 29.2 psi The unit conversion is as follows: 1 atm = 14.69 psi

1 atm P= 29.2 psí * 14.69 psí P=1.9877 atm P=1.99 atm Therefore, the inflated pressure, P is 1.99 atm. Calculate the inflated pressure (P in kPa) as shown below: P= 29.2 psi The unit conversion is as follows: 101.3 kPa = 14.69 psi P=29.2 pøí x 101.3 kPa P = 29.2 psi * 14.69 psí P=201.35 kPa P~201 kPa Therefore, the inflated pressure, P is 201 kPa.

Calculate the inflated pressure (P in in Hg) as shown below: P= 29.2 psi The unit conversion is as follows: 29.92 in Hg = 14.69 psi P= 29.2 psix 29.92 in Hg 14.69 ps1 P = 59.47 in Hg P = 59.5 in Hg Therefore, the inflated pressure, P is 59.5 in Hg|. Calculate the inflated pressure (P in mm Hg) as shown below: P= 29.2 psi The unit conversion is as follows: 760 mm Hg = 14.69 psi

P= 29.2 psí x 760 mm Hg 14.69 psí P=1510.68 mm Hg P=1.51068x10 mm Hg P~1.51x109 mm Hg Therefore, the inflated pressure, P is 1.51x10 mm Hg. Calculate the inflated pressure (P in torr) as shown below: P= 29.2 psi The unit conversion is as follows: 14.69 psi = 760 torr 760 torr P = 29.2 psix 14.69 psí P=1510.68 torr P=1.51068x10° torr P=1.51x10 torr

Therefore, the inflated pressure, P is 1.51x10 torr. Therefore, the inflated pressure of an automobile tire of 29.2 psi in different units of pressure is as follows: psi atm 29.2 1.99 kPa 201.35 in Hgmm Hg torr 59.5 1.51x10° 1.51x10 Answer:3 The density of liquid isopropyl iodide, Pisopropyl iodide is 1.70 $ mL The density of liquid mercury, Pmereury is 13.6 ml The unit conversions are as follows: 1 kg = 1000 g 1 mL = 10 m

The density of liquid isopropyl iodide in ky is as follows: Pisopropyl iodide mL 1000 g Pisopropyl iodide Pisopropyl iodide = 1.7x10-3 kg ml Pisopropyl iodide =1.7 x10-3 kg mit 1 m 10 m Pisopropyl iodide mº The density of liquid mercury in kg is as follows: mº

Pemercury = 13.6 x_1kg mL 1000 g Pmercury = 13.6x10-3 kg mL P mercury = 13.6x10- kg, 1 m mł 106 m P mercury = 13600 kg The atmospheric pressure, P is 0.964 atm. The unit conversions are as follows: 1 atm =1.013 x10ʻ Pa 1 Pa=1 kg m.s?

The atmospheric pressure in kg, is as follows: m.s? P=0.964 atin x 1.013 x10 Pa P=0.964 atm = 1 am P=97653.2 Pa kg 1 Da P=97653.2 Páx mois P=97653.2 kg m.s? The relation between the pressure (P) and the height of the fluid (liquid isopropyl iodide) in the barometer (h) is as follows: P = Pisopropyl iodidegh Here, g is the acceleration of gravity and its value is 9.8

Calculate the height of the fluid (liquid isopropyl iodide) in the barometer in cm as shown below: P= Pisopropyl iodide gn P ha Pisopropyl iodide 8 97653.2 h= Mist 1700 8x9,m h=5.86 m The unit conversion is as follows: 1 m = 100 cm h =5.86 x 100 cm IM h = 586 cm Therefore, the height of the fluid (liquid isopropyl iodide) in the barometer, h is 586 cm.