Ideal gas law problems solutions

    The gases are characterized by the lack of definite volume or shape. In the gaseous state, the matter has the properties of filling completely any available space to a uniform density.
    Low density and high compressibility are also pronounced characteristics of gases.
Ideal gas law problems solutions
Ideal gas laws
  • Problem
    Explain an ideal gas cannot be liquefied?
  • Solution
    For gas to liquefy, it is necessary for the molecules to come close to each other. For that, the gas should have the intermolecular attraction.
    But in an ideal gas, it is assumed that there is no intermolecular attraction and hence it can not be liquefied.
    Further putting the condition of the critical state of the ideal gas equation, PV = RT, it is possible to show that the critical temperature(Tc) for an ideal gas is 0⁰ K.
For ideal gas law,
or, (dP/dV)T = - RT/V²

At the critical state (dP/dV)T = 0

hence (- RTc/Vc²) = 0
or, Tc = 0⁰ K.
    Thus to liquefy an ideal gas, it is necessary to bring the temperature below 0⁰ K and it is not possible and so the ideal gas cannot be liquefied.
  • Problem
    Find the density of ammonia gas at 100⁰ C, when confined by the pressure of 1600 mm Hg.
  • Solution
The ideal gas equation is,
PV = nRT
or, PV = (g/M)RT
or, PV = (g/V)(RT/M)
or, P = dRT/M
or, d = PM/RT
In the above problem molecular weight M = 17 gm mol⁻¹,
Pressure P = 1600 mm Hg = 1600/760 atm,
R = 82.1 cc atm mol⁻¹ K⁻¹ and T = (273 +100) K

Thus density (d) = (17 × 1600)/(760 × 82.1 × 373)
= 0.00117 gm/cc
  • Problem
    Express the coefficient of thermal expansion(α) of a gas. Show that α depends on the temperature for an ideal gas.
  • Solution
abcdCoefficient of thermal expansion(α) is defined as, α = (1/V)[dV/dT]P

Ideal gas equation for 1 mole gas is,PV = RT
Hence [dV/dT]P = R/P
Thus α = (1/V) × (R/P) = (R/PV) = 1/T
This means all the gases have the same coefficient of thermal expansion.
  • Problem
    The compressibility factor Z of a gas is given by Z = PV/RT. What is the importance of this expression?
  • Solution
    The expression of Z provides the measures of non-ideality of a gas. It is the single parameter by the extent of non-ideality of gas could be measured.
    Thus when Z = 1, the gas behaves ideally.
    But when Z ≠ 1, the gas behaves non-ideal and departure of the value Z from one measures the extent of non-ideality of the gas.
When Z ㄑ1, PV ㄑRT
That is, PVㄑPi Vi
Thus the gas becomes more compressible than ideal gas.

Again when Z 〉1,
the gas is less compressible than the ideal gas.

  • Problem
    Show that Z is always greater than 1 for a gas obeying the equation P(V - b) = RT. What does the result signify?
  • Solution
The equation is given as,
P(V - b) = RT
or, PV = RT + bP
or, (PV/RT) = 1 + (Pb/RT)
or, Z = 1 + (Pb/RT)

From the above relation, Pb/RT is (+)ve thus Z 〉1.
  • Problem
    Why an ideal gas cannot have a reduced equation of state?
  • Solution
    The reduced equation of state is expressed in terms of reduced variables like reduced temperature (θ) reduced pressure (π) and reduced volume (φ).
These variables are defined as, θ = T/Tc
π = P/Pc
φ = V/Vc
where Tc, Pc, and Vc are the critical temperature, pressure, and volume respectively.

The ideal gas is represented by the equation,
PV = RT for 1 mole.
    But the critical temperature of the ideal gas is 0° K, hence the critical state is not attained by the ideal gas since 0°K is unattainable.
    Since critical Constants (Pc, Tc, and Vc) are not available for an ideal gas, so the reduced variables (π, θ, φ) are also not available for an ideal gas.
    Thus reduced equation of state cannot be constructed for an ideal gas.
  • Problem
    How the root means square velocity for oxygen compares with that of the hydrogen?
  • Solution
    The root means square velocity of hydrogen is four-times that of oxygen.
  • Problem
    Calculate the pressure excreted on the walls of a 3 liter of the flask when 7 gms of nitrogen are introduced into the same at 27°C.
  • Solution
We have, PV = nRT
or, PV = (g/M)RT (molecular wt. of N₂ = 28)
∴ P = (7/28) × (0.082 × 300)/3
= 2.05 atm

Here R = 0.082 lit atm mol⁻¹ K⁻¹
  • Problem
    What is the molecular weight of a gas, 12.8 gms of which occupy 10 liters at a pressure of 750 mm and at 27° C?
  • Solution
    31.91 gm mol⁻¹
  • Problem
    Calculate the total kinetic energy of 0.5 mol of an ideal gas at 273 K.
  • Solution
Total kinetic energy, = n (3RT/2)
= 0.05 {(3/2) × 8.314 × 273}

= 1702 J

Ideal gas law problems and solutions, Boyle's low, Charl's low, Combination of Boyle's Low and Charl's low, Ideal gas equation and kinetic theory of gases

[Chemical kinetics] [column1]

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