Calculation Of Molar Mass And Density Using The Ideal Gas Equation

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Calculation Of Molar Mass And Density Using The Ideal Gas Equation

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**

(a) One Of two 1 Liter flask contains a gas of molar mass 30, the other contains a gas of molar mass 60, both at the same temperature. The pressure in flask A is X atm, and the mass of gas in the flask is 1.2 g. The pressure in flask B is 0.5X atm, and the mass of gas in that flask is 1.2 g. Which flask contains gas of molar mass 30, and which flask contains the gas of molar mass 60?

(b) The *Hindenburg* was a hydrogen-filled dirigible that exploded in 1937. If the *Hindenburg* held 2 x 10^{5} m^{3} of hydrogen gas at 23^{o}C and 1 atm. What mass of hydrogen gas was present?

(c) Which gas is the most dense at 1 atm and 298^{o}K?

(i) CO_{2} (ii) N_{2}O (iii) Cl_{2}

(d) Why does a closed balloon filled with helium gas rise in air?

(e) Calculate the density of NO_{2} gas at 0.970 atm and 35^{o}C.

(f) Calculate the molar mass of a gas if 2.50 g occupies 0.875 L at 685 torr and 35^{o}C.

(g) The Dumas-bulb technique is used for determining the molar mass of an unknown by vaporizing the sample of liquid that boils below 100^{o}C in a boiling water bath and then determining the mass of vapor required to fill the bulb.

Calculate the molar mass of an unknown liquid, given that its vapor mass is 1.012 g; volume of bulb is 354 cm^{3}; pressure is 742 torr; and temperature is 99^{o}C.

**The strings**:
S_{7}P_{1}A_{15} (containership - mass).
**The math**:

Pj Problem of Interest is of type *containership* (mass).
**(a)** Let the number of moles of gas in flask A and flask B be n_{A}
and n_{B} respectively.

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

So, n_{A} = x(1)/(0.0821T) = x/(0.0821T).

Similarly, n_{B} = 0.5x/(0.0821T).

So, n_{A}/n_{B} = [x/(0.0821T)]/[0.5x/(0.0821T)] = 1/0.5

So, n_{B} = 0.5n_{A}.

1.2 g of gas with molar mass 30 = 1.2/30 = 0.04 mol

1.2 g of gas with molar mass 60 = 1.2/60 = 0.02 mol

Since 0.02 = 0.5(0.04), n_{B} = 0.02 and n_{A} = 0.04

So, flask A contains gas with molar mass 30 and flask B contains gas with molar mass 60.
**(b)** 1 L = 10^{-3} m^{3}

So, 2 x 10^{5} m^{3} = 10^{3}(2 x 10^{5}) = 2 x 10^{8} L

Mol of hydrogen gas present = n = PV/RT = (1 x 2 x 10^{8})/[(0.0821)(296)] = 8.23 x 10^{6} mols

Molar mass of H_{2} = 2

So, mass of 8.23 x 10^{6} mols = 2 x 8.23 x 10^{6} g = 16.5 x 10^{6} = 1.65 x 10^{4} kg.
**(c)** V = (nRT)/P (assuming n = 1 mole for each of the gases)

Volume of carbon dioxide = v_{CO2} = 1(0.0821)298/1 = 24.4658 L

Molar mass of carbon dioxide = m_{CO2} = 12 + 32 = 44 g

So, density of carbon dioxide = d_{CO2} = 44/24.4658 = 1.798 g/L.

Volume of N_{2}O = v_{N2O} = 1(0.0821)298/1 = 24.4658 L

Molar mass of N_{2}O = m_{N2O} = 14 x 2 + 16 = 44 g

So, density of N_{2}O = d_{N2O} = 44/24.4658 = 1.798 g/L.

Volume of Cl_{2} = v_{Cl2} = 1(0.0821)298/1 = 24.4658 L

Molar mass of Cl_{2} = m_{Cl2} = = 35.45 x 2 = 70.9 g

So, density of N_{2}O = d_{N2O} = 70.9/24.4658 = 2.898 g/L.

So, Chlorine is most dense at 1 atm and 298^{o}C.

In general, for gas samples at the same conditions, molar mass determines density
**(d)** Density of helium less than density of air. In other words, the balloon weighs less than the air displaced by its volume.
**(e)** V = nRT/p (assuming 1 mole of nitrogen dioxide, NO_{2})

So, V = 1(0.0821)(308)/0.970 = 26.069 L

Molar mass of NO_{2} = 14 + 32 = 46 g

So, density of N)_{2} = 46/26.069 = 1.765 g/L.
**(f)** n = PV/RT

685 torr = 685/760 = 0.901 atm

So, n = 0.901(0.875)/(0.0821 x 308) = 0.788/25.287 = 0.031 mol.

Mass of 0.031 mol = 2.5 g

So, mass of 1 mole = 2.5/0.0312 = 80.1 g
**(g)** The following illustration is the Dumas-Bulb Diagram

n = VP/RT.

V = 354 cm^{3} = 354 mL = 0.354 L.

P = 742 torr = 742/760 = 0.976 atm

T = 99^{0}C = 273 + 99 = 372^{0}K

So, V = (0.354)(0.976)/(0.0821 x 372) = 0.346/30.54 =0.01133 mol.

mass of 0.01133 mole vapor = 1.012 g

So, mass of 1 mole of vapor = 1.012/0.01133 = 89.32 g.

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The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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