Question

a) Write down the state equation of an ideal gas. This equation can be utilized to derive any one of the macroscopic thermodynamic properties, p1, T1, or V1, of an ideal gas in state “1” when the other two properties and the gas’ conditions at some initial state “0” are known.

b) What is the ideal gas law? Express the law separately in terms of the amount of substance, the gas’ molar mass, the gas’ molar volume, and the gas’ number of particles (4 equations!).

c) When we divide the universal gas constant by Avogadro’s constant we obtain a new constant. What is this constant’s name? Derive its value and proper unit.

d) State Newton’s Law of Cooling and name all properties in the corresponding equation.

e) For most thermodynamic problems we use some standard definitions for pressure and temperature to which we relate our calculations. A popular choice is a standard air pressure of 1atm = 101325Pa at sea level and a temperature of T = 288.15K (or ϑ = 25°C), BUT there are many other standards in use! What is the barometric elevation equation which allows you to approximate the air pressure at an arbitrary height (idealized!)?

Answer 1

(a) Write down the state equation of an ideal gas.

We know that, the equation of states which relates the P, V, T and n.

For an ideal gas, we have

P V = n R T

It is a combination of Boyle's and Charles's law.

Boyle's law : At constant temperature (T), P 1/V, thus PV = constant

Charles's law : At constant temperature (P), V T

(d) State Newton’s Law of Cooling.

Newton's law of cooling states that "the rate of heat loss of a body is directly proportional to the difference in temperatures between the body and its surroundings provided the temperature difference is small and the nature of radiating surface remains same".

(e) What is the barometric elevation equation which allows us to approximate the air pressure at an arbitrary height?

P = P₀ exp [- (Mg / RT) h]

where, P₀ = average atmospheric sea level pressure = 101325 Pa

M = molar mass of Earth's air = 0.02896 kg/mol

g = acceleration due to gravity = 9.8 m/s²

R = universal gas constant = 8.314 J/mol.K

T = standard temperature = 288.15 K

h = height above sea level = ?

then, we get

P = (101325 Pa) exp {- [(0.02896 kg/mol) (9.8 m/s²) / (8.314 J/mol.K) (288.15 K)] h}

P = (101325 Pa) exp (-0.00012 h) Pa

converting Pa into kPa :

P = (101.325) exp (-0.00012 h) kPa