Prequiz 13.1 - Ideal Gas Law .:. Go Up

The Formulas:

P = Pressure in Pascals (1 Pa = 1 N/m

^{2}) Gauge pressure (P_{G}) is the amount of pressure more than one atmosphere. P = P_{G}+ 1 ATM

F = Force in N

A = Area in m^{2}. A =LxWfor a rectangle, andprfor a circle.^{2}

P = the absolute pressure in Pa. Don't use gauge pressure. If you get pressure in psi, or any other unit, and or gauge, convert it to Pa absolute:

4.0 psi gauge = 4.0 psi + 14.7 psi = 18.7 psi absolute (because 1 atm in psi is 14.7) and finally, 18.7/14.7 = P/1.013x10^{5}, so P = 126,797.2789 Pa absolute.

Remember, 1 atm = 1.013x10^{5}Pa = 101.3 kPa = 14.7 psi = 760 Torr.

V = Volume in cubic meters. (m^{3}) There are 1000 liters in 1 m^{3}.

n = number of mols. n = mass/molar mass.

R =8.31JKT = Absolute temperature in KELVINS. T =^{-1}mol^{-1 }^{o}C + 273.15

This is not in the data packet - This works if everything is absolute. (i.e. not gauge pressure, not Celcius, not change in mols or change in volume, but absolute mols and volume)

The Prequiz:

(1 atm =
1.013x10^{5} Pa = 101.3 kPa = 14.7 psi = 760 Torr; 1 m^{3} =
1000 liters; p_{absolute} = p_{gauge} + 1 atm; )

**1. A
porthole in an airplane has a diameter of 37.0 cm. If there is a pressure
of 412 Torr on the outside of the window, and a pressure of 745 Torr inside,
what is the net force pushing out on the porthole? **(4770 N)

The area of the porthole is

pr=^{2}p(0.185 m)= 0.107521009 m^{2 }^{2}.The pressure difference between the inside and outside is 745 T - 412 T = 333 T.

Convert to Pa:

333/760 = P/1.013x10^{5}, P = 44,385.39474 PaUse to find the force:

P = 44,385.39474 Pa, A = 0.107521009 m^{2}, so F = 4772.362408 N =4770 N(None of the skill set ones is remotely as hard)

**2. Fred
has 1.65 mols of methane gas at 87.2 ^{o}C at 56.3 kPa (1 kPa = 1000
Pa). What is the volume it occupies? (.0878 m^{3})**

P = 56.3 x 10

^{3}Pa

V = ?

n = 1.65 mols

R =8.31JKT = 87.2 + 273.15^{-1}mol^{-1 }

**3.
Maryland has 217 grams of Neon (molar mass 20.1797 g/mol) gas in 519 liters
at gauge pressure of 6.97x10^{4} Pa. What must the
temperature be in Celsius? (1000 liters = 1 m^{3}) (720.
^{o}C)**

P = 6.97x10

^{4}Pa + 1.013x10^{5}Pa

V = 519/1000

n = 217/20.1797

R =8.31JKT = ? (Subtract 273.15 to get Celcius)^{-1}mol^{-1 }

**4. An
aerosol can is at an absolute pressure of 381 Boogalas when it is at 293 K.
If I put it in liquid nitrogen and lower its temperature to 77.0 K, what is
the new pressure in Boogalas? (1000 milli Boogalas = 1 Boogala) (Assume it
does not leak, and the volume remains constant) (100. Boogalas)**

P

_{1}= 381 Boogalas

V_{1}= Assumed Constant

n_{1}=Assumed Constant

T_{1}= 293 K

P_{2}= ??

V_{2}= Assumed Constant

n_{2}= Assumed Constant

T_{2}= 77.0 K

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__**5. A
Helium tank contains 3.42 kg of helium and is at a gauge pressure of 145 psi.
What will be the gauge pressure when you have released 1.13 kg of
helium? (92.2 psi)**

P

_{1}= 145 + 14.7 psi

V_{1}= Assumed Constant

n_{1}=3.42

T_{1}= Assumed Constant

P_{2}= ?? (subtract 14.7 to get gauge)

V_{2}= Assumed Constant

n_{2}= 3.42-1.13

T_{2}= Assumed Constant

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