1) i. Derive the ideal gas equation.
ii. Deduce values of gas cosntant 'R' in
a. SI unit in Joules
b. Lit- atm
Answers
215.4K−273=−57.4ºC(18)
EXAMPLE 4
What is the density of nitrogen gas ( N2 ) at 248.0 Torr and 18º C?
SOLUTION
Step 1: Write down your given information
P = 248.0 Torr
V = ?
n = ?
R = 0.0820574 L•atm•mol-1 K-1
T = 18º C
Step 2: Convert as necessary.
(248Torr)×1atm760Torr=0.3263atm(19)
18ºC+273=291K(20)
Step 3: This one is tricky. We need to manipulate the Ideal Gas Equation to incorporate density into the equation. *Write down all known equations:
PV=nRT(21)
ρ=mV(22)
where ρ =density, m=mass, V=Volume
m=M×n(23)
where m=mass, M=molar mass, n=moles
*Now take the density equation.
ρ=mV(24)
*Keeping in mind m=M×n ...replace (M×n) for mass within the density formula.
ρ=M×nV(25)
ρM=nV(26)
*Now manipulate the Ideal Gas Equation
PV=nRT
nV=PRT(27)
* (n/V) is in both equations.
nV=ρM(28)
nV=PRT(29)
*Now combine them please.
ρM=PRT(30)
*Isolate density.
ρ=PMRT(31)
Step 4: Now plug in the information you have.
ρ=PMRT(32)
ρ=(0.3263atm)(2∗14.01g/mol)(0.08206Latm/Kmol)(291K)(33)
ρ=0.3828g/L(34)
EXAMPLE 5
Find the volume, in mL, when 7.00 g of O2 and 1.50 g of Cl2 are mixed in a container with a pressure of 482 atm and at a temperature of 22º C.
SOLUTION
Step 1: Write down your given information
P = 482 atm
V = ?
n = ?
R = 0.0820574 L•atm•mol-1 K-1
T = 22º C + 273 = 295K
1.50g Cl2
7.00g O2
Step 2: Find the total moles of the mixed gases in order to use the Ideal Gas Equation.
ntotal=nO2+nCl2(35)
=[7.0gO2×1molO232.00gO2]+[1.5gCl2×1molCl270.905gCl2](36)
=0.2188molO2+0.0212molCl2(37)
=0.24mol(38)
Step 3: Now that you have moles, plug in your information in the Ideal Gas Equation.
V=nRTP(39)
V=(0.24mol)(0.08206Latm/Kmol)(295K)(482atm)(40)
V=0.0121L(41)
Step 4: Almost done! Now just convert the liters to milliliters.
0.0121L×1000ml1L=12.1mL(42)
EXAMPLE 6
A 3.00 L container is filled with Ne(g) at 770 mmHg at 27oC. A 0.633g sample of CO2 vapor is then added. What is the partial pressure of CO2 and Ne in atm? What is the total pressure in the container in atm?
SOLUTION
Step 1: Write down all given information, and convert as necessary.
Before:
P = 770mmHg --> 1.01 atm
V = 3.00L
nNe=?
T = 27oC --> 300K
Other Unknowns: nCO2 = ?
nCO2=0.633gCO2×1mol44g=0.0144molCO2(43)
Step 2: After writing down all your given information, find the unknown moles of Ne.
nNe=PVRT(44)
nNe=(1.01atm)(3.00L)(0.08206atmL/molK)(300K)(45)
nNe=0.123mol(46)
Because the pressure of the container before the CO2 was added contained only Ne , that is your partial pressure of Ne . After converting it to atm, you have already answered part of the question!
PNe=1.01atm(47)
Step 3: Now that have pressure for Ne, you must find the partial pressure for CO2 . Use the ideal gas equation.
PNeVnNeRT=PCO2VnCO2RT(48)
but because both gases share the same Volume ( V ) and Temperature ( T ) and since the Gas Constant ( R ) is constants, all three terms cancel and can be removed them from the equation.
PnNe=PnCO2(49)
1.01atm0.123molNe=PCO20.0144molCO2(50)
PCO2=0.118atm(51)
This is the partial pressure CO2 .
Step 4: Now find total pressure.
Ptotal=PNe+PCO2(52)
Ptotal=1.01atm+0.118atm(53)
Ptotal=1.128atm≈1.13atm(with appropriate significant figures)(54)
Answer:
(1)Let us write the equation of three gas laws:
1)Boyle's law:
Pressure is inversely proportional to Volume at constant Temp.
2)Charles law:
Volume is directly proportional to Temperature at Constant Pressure.
3)Avogadro law:
Volume is directly proportional to number of moles.
Combining all the 3 laws we get,
V¢nT/P
V=R nT/P
Where R is a proportionality constant or gas constant.
On rearranging above equation we get,
PV=nRT
This equation is known as Ideal Gas equation.
(2)R= 8.314 J/K/mol in Joules
R= 0.0821 L atm/K/mol in L atm