Calculation Of Average Kinetic Energy, Effusion Rate And Molar Mass Of A Gas

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Calculation Of Average Kinetic Energy, Effusion Rate And Molar Mass Of A Gas

Given that volume = V liters; pressure = P atmosphere; temperature = T Kelvin; Numuber of Moles = n.
**Boyle's Law: V α 1/P (constant n, T)**
**Charle's Law: V α T (constant n, P)**
**Avogadro's Law: V α n (constant P, T)**
**Ideal Gas Equation: V α nT/P ; V = nRT/p (R is proportionality constant = 0.0821 L-atm/mol-K**
**Graham's Law Of Effusion: the effusion rate of a gas is inversely proportional to the square root of its molar mass**.
**(a)** Vessel A contains CO_{2}(g) at 0^{o}C and 1 atm. Vessel B contains SO_{2}(g) at 20^{o}C and 0.5 atm. The two vessels have the same volume:

(i) Which vessel contains more molecules?

(ii) Which vessel contains more mass?

(iii) In which vessel is the average kinetic energy higher?

(iv) In which vessel is the rms speed of the molecules higher?
**(bi)** Place the following gases in order of increasing average molecular speed at 25^{o}C: Ne, HBr, SO_{2}, NF_{3}, CO.

(ii) Calculate the rms speed of NF_{3} molecules at 25^{0}C.
**(c)** Hydrogen has two naturally occurring isotopes.^{1}H and ^{2}H. Chlorine also has two naturally occurring isotopes, ^{35}Cl and ^{37}Cl. Thus hydrogen chloride consists of four distinct types of molecules: ^{1}H^{35}Cl, ^{1}H^{37}Cl, ^{2}H^{35}Cl, and ^{2}H^{37}Cl. Place these four molecules in order of increasing rate of effusion.
**(d)** A gas of unknown molecular mass was allowed to effuse through a small opening under constant pressure conditions. It required 105 seconds for 1 L of the gas to effuse. Under identical experimental conditions it required 31 seconds for 1 L O_{2} gas to effuse. Calculate the molar mass of the unknown gas.

**The strings**:
S_{7}P_{4}A_{41} (motion - linear)

S_{7}P_{1}A_{15} (containership - mass)

S_{7}P_{5}A_{51} (change - physical)
**The math**:

There are three Pj Problems of interest: kinectic energy which is energy due to motion(S_{7}P_{4}A_{41}), root mean speed which is due to change in speed (S_{7}P_{5}A_{51}), and mass of gas which is a containership problem (S_{7}P_{1}A_{15}).
**(ai)** Let the number of moles of gas in vessel A and vessel B be n_{A} and n_{B} respectively

From the ideal gas equation, n_{A} = P_{A}V_{A}/RT_{A}

So, n_{A} = n_{CO} = (1 x V)/273R.

Similarly, n_{B} = n_{SO2} = 0.5xV/(293R).

So, n_{A}/n_{B} = [(1 x V)/273R]/[0.5 x V/(293R)] = 2(293)/273 = 2.147

So, n_{A} = 2.147 n_{B}

So, vessel A contains more molecules.
**(aii)** Molar mass of CO = 12 + 16 = 28 g

So, mass of n_{A} mol of CO = m_{A} = 28n_{A} = 28(2.147n_{B}) =60.116n_{B} g

Molar mass of SO_{2} = 32 + 32 = 64 g

So, mass of n_{B} mol of SO_{2} = m_{B} = 64n_{B} g

So, vessel B contains more mass since m_{B} > m_{A}

This solution can also be arrived at by comparing density of CO (1.25 g/L) and density of SO_{2} (1.33 g/L).
**(aiii)** Average kinetic energy = 1/2(mu^{2})

Where u is the root mean speed = (3RT/M)^{1/2}

Where R is the constant 0.0821 L-atm/mol, T is temperature in degree Kelvin , and M is molar mass.

Average kinetic energy will be higer in vessel B which has greater mass molecules and are at higher temperature.
**(aiv)** Root mean speed (rms) = (3RT/M)^{1/2}

So, rms_{A} = u_{A} = [(3x8.314x273)/0.028]^{1/2} = (2.43 x 10^{5})^{1/2} m/s

So, rms_{B} = u_{B} = [(3x8.314x293)/0.064]^{1/2} = (1.14 x 10^{5})^{1/2} m/s

So, u_{A}/u_{B} = 1.46

So, u_{A} > u_{B}.
**(bi)** For an ideal gas, the average speed = 0.921(rms)

So, for gases at the same temperature, the gas with the least molar mass is the gas with the greatest molecular speed

molar masses:

Ne = 20.1797; HBR = 1 + 79.904 = 80.904; SO_{2} = 32 + 32 = 64; NF_{3} = 14 + 57 = 71; CO = 12 + 16 = 28.

So, in order of increasing molecular speed:

HBR < NF_{3} < SO_{2} < CO < Ne.
**(bii)** rms_{NF3} = u_{NF3} =[(3x8.314x298)/(71 x 10^{-3})]^{1/2}
Where R is Joule/mol-K = 8.314 J/mol-K; and molar mass is in kg.

So, u_{NF3} =[(3x8.314x298)/(71 x 10^{-3})]^{1/2} = (104686)^{1/2} = 323.55 m/s.
**(c)** Molar masses:
^{1}H^{35}Cl = 1 + 35 = 36
^{1}H^{37}Cl = 1 + 37 = 38
^{2}H^{35}Cl = 2 + 35 = 37
^{2}H^{37}Cl = 2 + 37 = 39

By graham's law, the least mass effuse the fastest

So, the order of increasing rate of effusion is:
^{2}H^{37}Cl < ^{1}H^{37}Cl < ^{2}H^{35}Cl <sup>1H^{35}Cl.
**(d)**

By Graham's law rate of effusion = (k/M)^{1/2}

Where k is a constant, M is molar mass of gas

Let rate of effusion per liter of unknown gas be r_{x}

Let rate of effusion per liter of O_{2} be r_{O2}.

Then, r_{x} = 1/105 L/s

And r_{O2} = 1/31 L/s

So, r_{O2}/r_{x} = (M_{x}/M_{O2})^{1/2}

So, (105/31)^{2} = M_{x}/M_{O2} = M_{x}/32

So, 32(3.387)^{2} = M_{x} = 32x11.47 = 367 g

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