Aerodynamics Questions and Answers – A Brief Review of Thermodynamics – 1

This set of Aerodynamics Multiple Choice Questions & Answers (MCQs) focuses on “A Brief Review of Thermodynamics – 1”.

1. The equation of state for a perfect gas is p = δRT, where R is______ (for air at standard conditions).
a) 287 kJ/kg.K
b) 8314 kJ/kg
c) 8.314 kJ/kgK
d) 287 J/kg.K
View Answer

Answer: d
Explanation: The universal gas constant R in the equation of gas has different values for different gases and at different conditions. For air at standard conditions, the correct value is 287 J/kg.K (check units).

2. Which is the wrong result for a calorically perfect gas?
a) e=cvT
b) h=cpT
c) cp and cp are functions of T
d) For T < 1000 K, specific heats are constant
View Answer

Answer: c
Explanation: The calorically perfect gases have constant specific heats at low temperatures (T < 1000 K). cp and cp are nothing else but specific heat constants at constant pressure and temperature respectively. Internal energy and enthalpy can be given in terms of the specific heats i.e. e=cv T and h=cpT.

3. Specific heat for gas is almost constant in the case of_______
a) Space vehicle
b) High-temperature flow
c) Air in a desert
d) Chemically reacting flow at high- speeds
View Answer

Answer: c
Explanation: Instances of very high temperature, chemically reacting flow at high-speeds do not have constant specific heats. Space vehicles fall in this category. The air in any desert would not have temperatures higher than 1000K and thus fall in the category of calorically perfect gas (i.e. constant specific heat).
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4. Which is not a state variable/function in the given options according to thermodynamics?
a) Temperature
b) Enthalpy
c) Functions dependent on path
d) Internal energy
View Answer

Answer: c
Explanation: By definition, state functions do not depend on the path and only depend on the initial and final state of the gas. Temperature, enthalpy and internal energy are all path independent i.e. state functions.

5. The incorrect formula for the specific heat of an ideal gas in terms of specific heat ratio of a gas and universal gas constant is_____
a) cp-cv=2R
b) \(\frac {c_p}{c_v}\)=γ
c) cv=\(\frac {R}{\gamma -1}\)
d) cp=\(\frac {\gamma R}{\gamma -1}\)
View Answer

Answer: a
Explanation: The specific heat ratio of a gas is defined as the ratio \(\frac {c_p}{c_v}\) and the difference in cp and cv is defined as the universal gas constant. From these, the following relations for the specific heat of an ideal gas are obtained cp=\(\frac {\gamma R}{\gamma -1}\) and cv=\(\frac {R}{\gamma -1}\).
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6. For air, at 35°C what is the total enthalpy? (cp=1005 \(\frac {J}{kgK}\)).
a) Zero
b) 309.54 J
c) 309.54 kJ
d) 8977 J
View Answer

Answer: d
Explanation: The specific enthalpy (h) is given as h=cpT where T is in Kelvin. This is the enthalpy per unit mass. For total enthalpy of the air, we need to multiply h by the molecular mass of air i.e. H=29h which gives total enthalpy equal to 8977 J, irrespective of the process.

7. The first law of thermodynamics is proved by experiments. Select the correct choice for it.
a) de is an exact differential
b) dq is an exact differential
c) dw is an exact differential
d) The relation has been proved theoretically
View Answer

Answer: a
Explanation: The first law in thermodynamics is an empirical relation and has been verified by experiments. It has never been violated. Exact differential depends only on final and initial points. de is an exact differential while dq and dw are not exact differentials.
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8. Identify the incorrect source for the corresponding variables of the first law.
a) Absorbed radiation: dq
b) Squeezing of volume: dw
c) Heating: de
d) Displacement of system boundary: dq
View Answer

Answer: d
Explanation: The change in dq or dw brings a change in de. Displacement of system boundary or squeezing of system volume are sources of dw. While heating and absorbing of radiation by mass in the system are sources of dq, therefore de.

9. Which is the process in which no heat exchange is there between system and surroundings and in which no dissipative phenomena occur?
a) Adiabatic process
b) Reversible process
c) Isentropic process
d) Work
View Answer

Answer: c
Explanation: A reversible process is one in which no dissipative phenomena occur. An adiabatic process is one in which no heat exchange is there between the system and the surroundings. A process that is both adiabatic and reversible is called an isentropic process.
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10. For a reversible process, which is not the correct statement?
a) No presence of viscosity
b) dw = -pdv
c) dq – pdv = de
d) Mass diffusion occurs
View Answer

Answer: d
Explanation: Reversible process has no dissipative effects i.e. viscosity and mass diffusion are absent. Also, work in a reversible process is given as dw = -pdv. Therefore, the first equation of thermodynamics can be written as dq – pdv = de.

Sanfoundry Global Education & Learning Series – Aerodynamics.

To practice all areas of Aerodynamics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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