Ideal Gas Law: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | Go up
- by Allie Nishitani, 2004

1. What is the volume of one mol of an ideal gas at standard temperature and pressure (STP)?  (P = 1.000 atm = 1.013x105 Pa, T = 0oC = 273 K)  What is the volume in liters (1 m3 = 1000 liters)

V=?

n=1

p=1.013 E5

T=273 K

pV = nRT

v=nRT/p=1*8.135*273/1.013 E5=2.24E-2 m^3

Convert m^3 into L:

1 m^3=1000 L

2.24E-2*1000=22.4 L

2. Anita Breke fills a large helium balloon with 2.18 grams of Helium gas.  How many mols of He is this?  What is the pressure in the balloon if the gas occupies a volume of 12.05 liters (1000 liters = 1 m3) at a temperature of 18.0 oC? (Beware the ides of Celsius!)  What is that pressure in atmospheres?  What is the gauge pressure in Pa and Atmospheres?  Pgauge = Pabsolute – 1 atm)

First you have to find how many mols there are:

Mass=2.18 g

Molecular mass of He=4.002602

Mol=mass/molecular mass=2.18/4.002602=.545 mols

Now that you know how many mols there are, you can solve the rest of the problem:

p=?

V=12.05 L

1 m^3=1000 L

1/1000=x/12.05

x=.01205 m^3

T=18ºC

18+273=281 K

n=.545 mols

pV = nRT

p=nRT/V=.545*8.315*291/.01205=1.09E5 Pa

Convert Pa into atm:

101.3 KPa=1.00 atm

101.3/1.09 E2=1.00/x

x=1.08 atm

Find the gauge pressure in Pa and atm:

Pgauge=Pabsolute-1 atm

(1 atm=101.3 KPa)

Pg=1.09 E2 KPa-101.3 Kpa=7.7 Kpa=8E3 Pa

Pg=1.08 atm-1 atm=.08 atm

3.A reaction vessel operates at 3.14 atmospheres.  What is that pressure in Pa?  If the vessel has a volume of  .113 m3, is at a temperature of 145 oC, and contains pure Nitrogen gas, how many mols of nitrogen gas does it contain?  How many grams of Nitrogen does it contain?  (None Of Fred’s Clients Bring Iron Hats – Nitrogen is a diatomic gas)

p=3.14 atm

1 atm=101.3 KPa

1/3.14=101.3/x

x=318.082 KPa=3.18E5 Pa

V=.113 m^3

T=145 ºC

145+273=418 K

n=?

pV = nRT

n=pV/RT=3.18E5*.113/(8.315*418)=10.3 mol

Mol=mass/molecular mass

10.3=m/14.00674*2  (Nitrogen is diatomic)

M=290.0 g

4. A container has a volume of 216 liters.  (1000 liters = 1 m3)   If it can sustain a pressure of 13.5 atmospheres before bursting, and contains 89.1 grams of Hydrogen gas, a) what is its bursting pressure in Pa?  b) how many mols of Hydrogen does it contain?  and c) what is its maximum operating temperature in K and oC?

V=216 L

1 m^3=1000 L

1000/216=1/x

x=.216 m^3

p=13.5 atm

1 atm=101.3 KPa

1/13.5=101.3/x

x=1367550=1.37E6 Pa

n=mass/molecular mass=89.1/1.00794*2 (hydrogen is diatomic)=44.2 mols

T=?

pV = nRT

T=pV/nR=1.37E6*.216/(44.2*8.315)=804 K

Convert to Celsius:

804-273=531ºC

5. A 2.00 liter bottle contains 18.15 grams of Bromine gas and is at a gauge pressure of .153 atm.  What is its temperature in Celsius?

V=2 L

1 m^3=1000 L

1000/2=1/x

x= .002 m^3

n= mass/molecular mass=18.15/79.914=.11357 mol

P=Pg+Pa=.153+1=1.153 atm

1 atm=101.3 KPa

1/1.153=101.3E3/x

x=116798.8 Pa

pV = nRT

T=pV/nR=116798.8*.002/(.11357*8.315)=247.3676 K

Convert to Celsius:

247.3676 -273=-26ºC

6. A Tupperware container is at 1.00 atm at 21.0 oC.  (Convert to K)  It is heated in a microwave to 99.5 oC with the lid on.  Assuming no gas escapes, what is the pressure inside in atm?

p1=1

p2=?

T1=21ºC

21+273=294 K

T2=99.5 ºC

99.5+273=372.5 K

p1*V1/T1=p2*V2/T2

p2=1*372.5/294=1.27 atm

7. A quantity of ideal gas is compressed at constant temperature from 34.5 liters to 12.4 liters.  What was the initial pressure if the final pressure was 2.45x105 Pa?

V1=34.5 L

V2=12.4 L

p1=?

p2=2.45E5 Pa

p1*V1/T1=p2*V2/T2

p1=p2*V2/V1=2.45E5*12.4/34.5=8.81E4 Pa

8. A balloon has a volume of 1.25 liters at 20.5 oC.  At what temperature does it have a volume of 1.02 liters, assuming the pressure and mols remain constant?  Find the temperature in K and oC

V1=1.25 L

V2=1.02 L

T1=20.5 °C

20.5+273=293.5 K

T2=?

p1*V1/T1=p2*V2/T2

T2=V2*T1/V1=1.02*293.5/1.25=239 K

Convert to Celsius:

239-273= -34 °C

9. One mol of an ideal gas occupies 22.4 liters at STP.  (P = 1.000 atm  T = 0.00oC)  What volume does it occupy at 97.0 oC and 1.29 atm?

V1=22.4 L

p1=1 atm

T1=273 K

V2=?

T2=97

97+273=370 K

p2=1.29 atm

p1*V1/T1=p2*V2/T2

V2=p1*V1*T2/p2*T1=1*22.4*370/(1.29*273)=23.5 L