Ideal and Real Gases MCQ Quiz - Objective Question with Answer for Ideal and Real Gases - Download Free PDF

Concept:

For isentropic flow of an ideal gas, the ratio of stagnation temperature T₀ to critical temperature T* can be determined using the specific heat ratio γ. The critical temperature is the temperature at the throat (Mach number = 1) in a nozzle.

Given:

• Specific heat ratio, \(γ = 1.289\)

Step 1: Recall the Stagnation-to-Critical Temperature Ratio Formula

The ratio is given by:

\[ \frac{T_0}{T^*} = \frac{γ + 1}{2} \]

This formula applies when the Mach number M is 1 (critical condition).

Step 2: Substitute the Given Value of γ

\[ \frac{T_0}{T^*} = \frac{1.289 + 1}{2} = \frac{2.289}{2} = 1.1445 \]

Concept:

From the ideal gas law: \( P = \rho R T \Rightarrow R = \frac{P}{\rho T} \)

Weight density \( w = \rho g \Rightarrow \rho = \frac{w}{g} \)

Calculation:

\( w = 20~\text{N/m}^3, \, g = 9.81~\text{m/s}^2 \Rightarrow \rho = \frac{20}{9.81} = 2.039~\text{kg/m}^3 \)

\( P = 0.26~\text{N/mm}^2 = 0.26 \times 10^6~\text{N/m}^2 \), \( T = 27^\circ C = 300~K \)

\( R = \frac{260000}{2.039 \times 300} = \frac{260000}{611.7} \approx 425.1~\text{J/kg-K} = 0.4251~\text{kJ/kg-K} \)

\(R=\cfrac {\bar R}{M}\)

where,

\(\bar R\)= Universal gas constant

\(R\)= specific gas constant and

\(\bar R=RM\)

\(\because C_P-C_V=\bar R\) & \(\cfrac {C_P}{C_V}=\gamma\)

\(\therefore C_V=\cfrac {C_P}{\gamma}\)

\(\therefore C_P-\cfrac {C_P}{\gamma}=\bar R\)

\(\therefore \left(\cfrac {\gamma-1}{\gamma}\right)C_P=\bar R= RM\)

\(\therefore C_P=\cfrac {\gamma RM}{\gamma-1}\)

Hence, option (A) is the correct answer.

Explanation:​

Real gases:

• The ideal gas law is only an approximation to the actual behaviour of the gases.
• At high densities that is at high pressure and low temperature the behaviour of the actual or real gases deviate from that predicted by the ideal gas law.
• In general at sufficiently low pressure or at low densities all gases behave like ideal gases.
• The ideal gases follow the characteristic gas equation and real gases follow the Van der Waals equation of state.

The ideal gas equation neglects the intermolecular forces and the volume occupied by the gas molecules. Vanderwall's gas equation takes into account these effects. Therefore real gasses follow Van der Waal's gas equation 

​\(\left( {P + \frac{a}{{{V^2}}}} \right)\left( {v - b} \right) = RT\)​

The constants a and b have positive values and are characteristic of the individual gas. The van der Waals equation of state approaches the ideal gas equation when values of these constants approach zero.

• The constant a provides a correction for the intermolecular forces (force of cohesion)
• Constant b is a correction for finite molecular size and its value is the volume of one mole of the atoms or molecules.​

Calculation:

Given:

The curve shows that V ∝ 1/T

⟹ V = m (1/T), where m = slope of the curve

⟹ V × T = constant

Differentiating both sides:

T dV + V dT = 0 ........(i)

⟹ dV = - (V / T) dT

From the first law of thermodynamics:

Q = ΔU + W

CdT = C_V dT + p dV ........(ii)

Substituting dV from equation (i):

CdT = C_V dT - (p / T) V dT

⟹ C = C_V - (pV / T)

Using the ideal gas equation, pV = RT:

⟹ C = C_V - R

For a **diatomic gas**, the molar heat capacity at constant volume:

C_V = (5/2) R

⟹ C = (5/2) R - R

= (3/2) R

∴ The heat capacity C is (3/2) R.

Concept:

Compressibility factor Z  is given by

\(Z = \frac{{PV}}{{RT}}\)

Where P is pressure, V is the molar volume of the gas, R is Universal gas constant, T is temperature.

At critical point:

\(P = \frac{a}{{27{b^2}}}\)

V = 3b

\(T= \frac{8a}{{ 27{R b }}}\)

Calculation:

Given:

\(Z = \frac{{PV}}{{RT}}\)

Putting the value of P, V, T in Z.

\(Z = \frac{{\frac{a}{{27b^2}}\times\;3b}}{{R\;\times\;\frac{8a}{27Rb}}}=\frac{3}{8}=0.375\)

Explanation:

If steam is throttled, its enthalpy remains constant and pressure drop takes place.

Throttling (Isenthalpic) process:

• A throttling process occurs when a fluid flowing through a passage suddenly encounters a restriction in the passage.
• The restriction could be due to the presence of an almost completely closed valve or due to sudden and large reduction in flow area etc.
• The result of this restriction is a sudden drop in the pressure of the fluid as it is forced to flow through the restriction. This is a highly irreversible process and is used to reduce the pressure and temperature of the refrigerant in a refrigeration system.
• Since generally, throttling occurs in a small area, it may be considered as an adiabatic process (as the area available for heat transfer is negligibly small) (δQ = 0) also since no external work is done (W = 0).

∴ h1 = h2

Explanation:

All vapour at a temperature higher than the critical temperature T_c is considered as gases (highly super-heated vapour).

• When a vapour condenses from a temperature below than the critical temperature T_c it changes it states from vapour to liquid.
• When gases (highly super-heated vapour) condenses, it falls into the super-critical state which is not easily liquifiable.
• A gas (highly superheated vapour) is thus approximated as an ideal gas as it never comes under liquid phase.

Concept:

Ideal gas equation:

PV = mRT

P = ρRT

\(ρ = \frac{P}{{RT}}\)

Mass of the air leaving the exit valve per unit time is given by:

ṁ = ρAV

ρ is the density of the air, A is the area of the exit valve and V is the exit velocity of the air.

Calculation:

Given:

At the exit of compressor

ṁ = 0.8 kg/s, P = 10 bar = 1000 kPa, T = 90°C = 363 K, Area = 2 × 10^–3 m²

From the equation, P = ρRT

1000 = ρ × 0.287 × 363

ρ = 9.59 kg/m³

Now, ṁ = ρ × A × V

0.8 = 9.59 × 2 × 10^-3 × V

V = 41.71 m/s.

Do not find the density at the inlet as the density is different. 

Concept:

The ideal gas equation is given by

PV = mRT

\(\Rightarrow P = \frac{m}{V}RT = \rho RT\)

P = ρRT 

For isobaric process

P₁ = P₂

∴ ρT = Constant, which is the equation for rectangular hyperbola. 

Explanation:

Joule–Thomson coefficient:

When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature. From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as

\(\mu = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H}\)

As a measure of the change in temperature which results from a drop in pressure across the construction.

• For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy.
• If μ is +ve, then the temperature will fall during throttling.
• If μ is -ve, then the temperature will rise during throttling.

Specific heat: It is defined as the energy required to raise the temperature of a unit mass of substance by one degree. In general this energy depend on how the process is executed.

In thermodynamics we deal with two kinds of specific heats

Specific heat at constant volume: It is the amount of energy required to raise the temperature of the substance by one degree if the volume is maintained constant.

Specific heat at constant pressure: It is the amount of heat required to raise the temperature of the substance by one degree if the temperature of the substance is raised by one degree if the pressure is held constant.

Specific heat for ideal gases

Monoatomic gas: The specific heat of monoatomic gases remain constant over the entire temperature range. 

Diatomic or triatomic gases: The specific heats of diatomic and triatomic gases increase with the increase in temperature. This variation is approximated as linear variation over small temperature intervals. Therefore the specific heats can be assumed as constant average specific heat values. 

Explanation:

Internal energy: U = u (T, V)

\(dU = {\left( {\frac{{\partial U}}{{\partial T}}} \right)_V} + {\left( {\frac{{\partial U}}{{\partial V}}} \right)_T}\;\)

For a real gas, the internal energy if a function of both the temperature and the specific volume.

That is,

\(U = f\left( {T,v} \right)\)

For ideal gas (∂U/∂V)_T = 0

∴ U = U (T) only

For an ideal gas (no intermolecular interactions and no molecular volume), appropriate equation of state would be: PV = nRT ⇒ V = nRT/P i.e. V = f(T,P,n)

There are many equations of state describing real gases. These equations take into consideration molecular volume and interactions. The most well-known such equation is the Van der Waals equation.

Explanation:

Joule – Thomson coefficient:

When the gas in steady flow passes through a constriction, e.g. in an orifice or valve, it normally experiences a change in temperature. From the first law of thermodynamics, such a process is isenthalpic and one can usefully define a Joule – Thomson coefficient as

\(\mu = {\left( {\frac{{\partial T}}{{\partial P}}} \right)_H}\)

As a measure of the change in temperature which results from a drop in pressure across the construction.

• For an ideal gas, μ = 0, because ideal gases neither warm not cool upon being expanded at constant enthalpy.
• If μ is +ve, then the temperature will fall during throttling.
• If μ is -ve, then the temperature will rise during throttling.

Explanation:

Vander-Waals equation

• The equation is basically a modified version of the Ideal Gas Law which states that gases consist of point masses that undergo perfectly elastic collisions. Ideal gas equation fails to explain the behaviour of real gases. Therefore, the Van der Waals equation was devised and it helps us define the physical state of a real gas.
• Van der Waals equation is an equation relating the relationship between the pressure, volume, temperature, and amount of real gases. For a real gas containing ‘n’ moles, the equation is written as

\(\left( {p + \frac{a}{{{v^2}}}} \right)\left( {v - b} \right) = RT\)

• The constant "a" provides a correction for the intermolecular forces.
• Constant b adjusts for the volume occupied by the gas particles. It is a correction for finite molecular size and its value is the volume of one mole of the atoms or molecules.

here, p is pressure in N/m², v is the specific volume in m³/kg, R is characteristic gas constant in J/kg-K, \( \frac{a}{{{v^2}}}\) is called the force of cohesion and b is called the co-volume.

(\(p + \frac{a}{{{v^2}}}\)) both terms should give the same unit since they are getting added

i.e. p = \( \frac{a}{{{v^2}}}\) in terms of unit.

\(\Rightarrow \frac{N}{{{m^2}}} = a{\left( {\frac{{kg}}{{{m^3}}}} \right)^2}\)

\(\Rightarrow a\left( {unit} \right) = \frac{{{m^6}}}{{k{g^2}}}.kg.\frac{m}{{{s^2}{m^2}\;}} = \frac{{{m^5}kg}}{{k{g^2}{s^2}}} = \frac{{{m^5}}}{{kg{s^2}}}\)