Properties of Gasses and Gas Mixture - 2 Free MCQ Practice Test with Solutions

Test: Properties of Gasses & Gas Mixture - 2 - Question 1

A football was inflated to a gauge pressure of 1 bar when the ambient temperature was 15°C. When the game started next day, the air temperature at the stadium was 5°C. Assume that the volume of the football remains constant at 2500 cm³.

The amount of heat lost by the air in the football and the gauge pressure of air in the football at the stadium respectively equal

A.
30.6 J, 1.94 bar
B.
21.8 J, 0.93 bar
C.
43.7 J, 0.93 bar
D.
61.1 J, 1.94 bar

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 1

Given:
Initial gauge pressure, P₁ = 1 bar
Initial temperature, T₁ = 15°C = 288 K
Final temperature, T₂ = 5°C = 278 K
Volume of football, V = 2500 cm³ = 2.5 × 10⁻³ m³
Volume constant
Gauge pressure means: P_gauge = P_absolute − P_atm
Atmospheric pressure P_atm = 1 bar = 10⁵ Pa
Air inside is treated as ideal gas
R = 8.314 J/mol·K
Molar mass of air M = 29 g/mol

Step 1: Find absolute initial pressure P₁ absolute
P₁(absolute) = P_atm + P_gauge = 1 + 1 = 2 bar = 2 × 10⁵ Pa

Step 2: Use ideal gas law to find number of moles n:
P V = n R T ⇒ n = (P V) / (R T)
n = (2 × 10⁵ × 2.5 × 10⁻³) / (8.314 × 288) = 500 / 2395.6 ≈ 0.2088 mol

Step 3: Find final pressure P₂ (absolute):
Since volume and amount of gas n are constant,
P₁ / T₁ = P₂ / T₂ ⇒ P₂ = P₁ × (T₂ / T₁) = 2 × 10⁵ × (278 / 288) = 1.93 × 10⁵ Pa

Step 4: Calculate final gauge pressure:
P_gauge,2 = P₂ − P_atm = 1.93 × 10⁵ − 1 × 10⁵ = 0.93 × 10⁵ Pa = 0.93 bar

Step 5: Calculate heat lost by the air (assuming air as ideal diatomic gas, C_V = (5/2) R):
Change in internal energy:
Q = n C_V ΔT = n × (5/2) × R × (T₂ − T₁)
Q = 0.2088 × 2.5 × 8.314 × (278 − 288)
Q = 0.2088 × 20.785 × (−10)
Q = 0.2088 × (−207.85) = −43.4 J

The negative sign indicates heat is lost.

Final answers:
Heat lost ≈ 43.7 J
Gauge pressure at stadium ≈ 0.93 bar

Correct option is:
c) 43.7 J, 0.93 bar

Test: Properties of Gasses & Gas Mixture - 2 - Question 2

A 100 W electric bulb was switched on in a 2.5 m × 3 m × 3 m size thermally insulated room having a temperature of 20°C. The room temperature at the end of 24 hours will be

A.
321°C
B.
341°C
C.
450°C
D.
470°C

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 2

Heat produced by electric bulb in 24 hr.

Test: Properties of Gasses & Gas Mixture - 2 - Question 3

2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in a mixing chamber, so that the final total pressure and temperature of the mixture become same as those of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing, per mole of oxygen, is given by

A.
–Rln2
B.
0
C.
Rln2
D.
Rln4

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 3

Remember if we mix 2 mole of oxygen with another 2 mole of other gas the volume will be doubled for first and second constituents

Total Entropy change = 4Rln2 So, Entropy change per mole=Rln2. And it is due to diffusion of one gas into another.

Test: Properties of Gasses & Gas Mixture - 2 - Question 4

Assertion (A): A perfect gas is one that satisfies the equation of state and whose specific heats are constant.

Reason (R): The enthalpy and internal energy of a perfect gas are functions of temperature only.

A.
Both A and R are individually true and R is the correct explanation of A
B.
Both A and R are individually true but R is not the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 4

For perfect gas, both the assertion A and reason R are true. However R is not the explanation for A. A provides definition of perfect gas. R provides further relationship for enthalpy and internal energy but can't be reason for definition of perfect gas.

Test: Properties of Gasses & Gas Mixture - 2 - Question 5

Consider an ideal gas contained in vessel. If intermolecular interaction suddenly begins to act, which of the following happens?

A.
The pressure increase
B.
The pressure remains unchanged
C.
The gas collapses
D.
None of these

Test: Properties of Gasses & Gas Mixture - 2 - Question 6

Consider the following statements:

A real gas obeys perfect gas law at a very

1. High temperature

2. High-pressure

3. Low pressure

Which of the following statements is/are correct?

A.
1 alone
B.
1 and 3
C.
2 alone
D.
3 alone

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 6

In Perfect gas intermolecular attraction is zero. It will be only possible when intermolecular distance will be too high. High temperature or low pressure or both cause high intermolecular distance so choice 1 and 3.

Test: Properties of Gasses & Gas Mixture - 2 - Question 7

If a real gas obeys the Clausius equation of state p(v – b) = RT then,

A.
B.
C.
D.

Test: Properties of Gasses & Gas Mixture - 2 - Question 8

Reduced pressure is

A.
Always less than atmospheric pressure
B.
Always unity
C.
An index of molecular position of a gas
D.
Dimensionless

Test: Properties of Gasses & Gas Mixture - 2 - Question 9

The internal energy of a gas obeying Van der Waal‟s equation

A.
Temperature
B.
Temperature and pressure
C.
Temperature and specific volume
D.
Pressure and specific volume

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 9

Joule‟s law states that for an Ideal gas internal energy is a function of temperature only. u = f(T). But this is not Ideal gas it is real gas.

Test: Properties of Gasses & Gas Mixture - 2 - Question 10

Which one of the following statements is correct?

A.
Compressibility factor is unity for ideal gases
B.
Compressibility factor is zero for ideal gases
C.
Compressibility factor is lesser than unity for ideal gases
D.
Compressibility factor is more than unity for ideal gases

Test: Properties of Gasses & Gas Mixture - 2 - Question 11

The value of compressibility factor for an ideal gas may be:

1. less or more than one

2. equal to one

3. zero 4. less than zero

The correct value(s) is/are given by:

A.
1 and 2
B.
1 and 4
C.
2 only
D.
1 only

Test: Properties of Gasses & Gas Mixture - 2 - Question 12

Assertion (A): An adiabatic process is always a constant entropy process.

Reason(R): In an adiabatic process there is no heat transfer.

A.
Both A and R are individually true and R is the correct explanation of A
B.
Both A and R are individually true but R is not the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Test: Properties of Gasses & Gas Mixture - 2 - Question 13

In which of the following processes, the heat is fully converted into work?

A.
Reversible adiabatic process
B.
Reversible isobaric process
C.
reversible isometric process
D.
Reversible isothermal process

Test: Properties of Gasses & Gas Mixture - 2 - Question 14

An ideal gas undergoes an isothermal expansion from state R to state S in a turbine as shown in the diagram given below: The area of shaded region is 1000 Nm. What is the amount is turbine work done during the process?

A.
14,000 Nm
B.
12,000 Nm
C.
11,000Nm
D.
10,000Nm

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 14

Turbine work = area under curve R–S

Test: Properties of Gasses & Gas Mixture - 2 - Question 15

The slope of log P-log V graph for a gas for isothermal change is m₁ and for adiabatic changes is m₂. If the gas is diatomic gas, then

A.
m₁<m₂
B.
m₁>m₂
C.
m₁ + m₂ = 1.0
D.
m₁ = m₂

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 15

PV = constant, C

Test: Properties of Gasses & Gas Mixture - 2 - Question 16

Air is being forced by the bicycle pump into a tyre against a pressure of 4-5 bars. A slow downward movement of the piston can be approximated as

A.
Isobaric process
B.
Adiabatic process
C.
Throttling process
D.
Isothermal process

Test: Properties of Gasses & Gas Mixture - 2 - Question 17

The change in specific entropy of a system undergoing a reversible process is given by This is valid for which one of the following?

A.
Adiabatic process undergone by an ideal gas
B.
Isothermal process undergone by an ideal gas
C.
Polytropic process undergone by a real gas
D.
Isobaric phase change from liquid to vapour

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 17

Integrating the above expression

For isothermal process undergone by ideal gas.

Test: Properties of Gasses & Gas Mixture - 2 - Question 18

Change in enthalpy in a closed system is equal to the heat transferred, if the reversible process takes place at

A.
Temperature
B.
Internal energy
C.
Pressure
D.
Entropy

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 18

Test: Properties of Gasses & Gas Mixture - 2 - Question 19

A saturated vapour is compressed to half its volume without changing its temperature. The result is that:

A.
All the vapour condenses to liquid
B.
Some of the liquid evaporates and the pressure does not change
C.
The pressure is double its initial value
D.
Some of the vapour condenses and the pressure does not change

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 19

By compressing a saturated vapour, its vapours condense and pressure remains unchanged. Remember it is not gas.

Test: Properties of Gasses & Gas Mixture - 2 - Question 20

Which one of the following thermodynamic processes approximates the steaming of food in a pressure cooker?

A.
Isenthalpic
B.
Isobaric
C.
Isothermal
D.
Isochoric

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 20

Since the volume of pressure cooker is constant hence it is an isochoric process.

Test: Properties of Gasses & Gas Mixture - 2 - Question 21

A system at a given state undergoes change through the following expansion processes to reach the same final volume

1. Isothermal

2. Isobaric

3. Adiabatic ( γ = 1.4)

4. Polytropic(n =1.3)

The correct ascending order of the work output in these four processes is

A.
3, 4, 1, 2
B.
1, 4, 3, 2
C.
4, 1, 3, 2
D.
4, 1, 2, 3

Test: Properties of Gasses & Gas Mixture - 2 - Question 22

Four processes of a thermodynamic cycle are shown above in Fig.I on the T-s plane in the sequence 1-2-3-4. The corresponding correct sequence of these processes in the p- V plane as shown above in Fig. II will be

A.
C–D–A–B
B.
D–A–B–C
C.
A–B–C–D
D.
B–C–D–A

Test: Properties of Gasses & Gas Mixture - 2 - Question 23

A perfect gas at 27°C was heated until its volume was doubled using the following three different processes separately

1. Constant pressure process

2 Isothermal process

3. Isentropic process

Which one of the following is the correct sequence in the order of increasing value of the final temperature of the gas reached by using the above three different processes?

A.
1 – 2 – 3
B.
2 – 3 – 1
C.
3 – 2 – 1
D.
3 – 1 – 2

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 23

Test: Properties of Gasses & Gas Mixture - 2 - Question 24

Assertion (A): For a perfect gas, hyperbolic expansion is an isothermal expansion.

Reason (R): For a perfect gas, PV/T = constant.

A.
Both A and R are individually true and R is the correct explanation of A
B.
Both A and R are individually true but R is not the correct explanation of A
C.
A is true but R is false
D.
A is false but R is true

Test: Properties of Gasses & Gas Mixture - 2 - Question 25

An ideal gas with initial volume, pressure and temperature of 0.1 m³, 1bar and 27°C respectively is compressed in a cylinder by a piston such that its final volume and pressure are 0.04 m3 and 5 bars respectively, then its final temperature will be:

A.
–123°C
B.
54°C
C.
327°C
D.
600°C

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 25

Test: Properties of Gasses & Gas Mixture - 2 - Question 26

An Ideal gas with initial volume, pressure and temperature of 0.1m³, 1 bar and 27°C respectively is compressed in a cylinder by piston such that its final volume and pressure 0.04 m3 and 5 bar respectively, then its final temperature will be:

A.
–123°C
B.
54°C
C.
327°C
D.
600°C

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 26

Apply equation of states

Test: Properties of Gasses & Gas Mixture - 2 - Question 27

If a gas obeys van der Waals' equation at the critical point, then is equal to which one of the following

A.
0
B.
1
C.
1·5
D.
2·67

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 27

Test: Properties of Gasses & Gas Mixture - 2 - Question 28

A higher value of Van der waal‟s constant for a gas indicates that the

A.
Molecules of the gas have smaller diameter
B.
Gas can be easily liquefied
C.
Gas has higher molecular weight
D.
Gas has lower molecular weight

Test: Properties of Gasses & Gas Mixture - 2 - Question 29

Consider the following statements:

1. During a reversible non-flow process, for the same expansion ratio, work done by a gas diminishes as the value of n in pvⁿ = C increases.

2. Adiabatic mixing process is a reversible process.

Which of the statements given above is/are correct?

A.
1 only
B.
2 only
C.
Both 1 and 2
D.
Neither 1 nor 2

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 29

In adiabatic mixing there is always
increase in entropy so large amount of irreversibility is these.

Test: Properties of Gasses & Gas Mixture - 2 - Question 30

In a reversible isothermal expansion process, the fluid expands from 10 bar and 2 m³ to 2 bar and 10 m³. During the process the heat supplied is at the rate of 100 kW. What is the rate of work done during the process?

A.
20 kW
B.
35 kW
C.
80 kW
D.
100 kW

Detailed Solution for Test: Properties of Gasses & Gas Mixture - 2 - Question 30

For reversible isothermal expansion heat supplied is equal to work done during the process and equal to

∵ Temperature constant so no change in internal energy dQ = dU + dW; dU = 0 Therefore dQ = dW.

Test: Properties of Gasses & Gas Mixture - 2 - Mechanical Engineering MCQ

30 Questions MCQ Test GATE Mechanical (ME) Mock Test Series 2026 - Test: Properties of Gasses & Gas Mixture - 2

A 100 W electric bulb was switched on in a 2.5 m × 3 m × 3 m size thermally insulated room having a temperature of 20°C. The room temperature at the end of 24 hours will be

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