This set of Gas Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Thermal and Calorical Properties and the Perfect Gas”.

1. A process operating at a constant pressure 2.3 bar and temperature of 650 K, cool down to a temperature of 288 K. Calculate the heat transfer and take specific heat at constant pressure 1005 J/Kg*K.

a) -363.81 KJ

b) -321.56 KJ

c) 363. 81 KJ

d) 321.56 KJ

View Answer

Explanation: The internal energy of the system in terms of temperature and pressure is written as:

h = h(T, p)

In terms of the exact differential equation; dh = \((\frac {\partial h}{\partial T})\)

_{p}dT + \((\frac {\partial h}{\partial p})\)

_{T}dp

For a constant pressure process dh = \((\frac {\partial h}{\partial T})\)

_{p}dT

dh = C

_{p}dT

Hence heat transfer Q = 1005 * (288 – 650)

Q = -363.81 KJ

2. The perfect gas must be both calorically and thermally perfect.

a) True

b) False

View Answer

Explanation: For a perfect gas both C

_{p}and C

_{v}are constant and are independent of temperature. Now for a calorically perfect gas, both C

_{p}and C

_{v}are constant and for a thermally perfect gas C

_{p}and C

_{v}are dependent on temperature, but their ratio is constant. Hence a perfect gas must be both calorically and thermally perfect.

3. The ratio of specific heats for helium is given by __________

a) 1.08

b) 1.24

c) 1.4

d) 1.67

View Answer

Explanation: For any simple molecular model, the ratio of specific heats is given by:

γ = (n + 2) / n

Where n = degree of freedom of molecules

And since helium is a monatomic gas, n = 3, thus γ= 1.67.

4. What will be the change in enthalpy for an adiabatic process?

a) dh = vdp

b) dh = pdv

c) dh = vdp + pdv

d) dh = dq + du

View Answer

Explanation: Since from the first law of thermodynamics

du = dq – dw

du = dq – pdv (as dw = pdv)

And since, h = u + pv

dh = du + pdv + vdp

Thus, dh = dq + vdp

Now for adiabatic pressure dq = 0, thus dh = vdp

5. Calculate the specific heat at constant pressure for a diatomic gas. Universal gas constant R = 287 J/Kg*K.

a) 718 J/Kg*K

b) 715 J/Kg*K

c) 1005 J/Kg*K

d) 428 J/Kg*K

View Answer

Explanation: Universal gas constant R = C

_{p}– C

_{v}

R / C

_{p}= 1 – C

_{v}/ C

_{p}

R / C

_{p}= (γ – 1) / γ (Since, γ= C

_{p}/ C

_{v})

C

_{p}= γR / (γ – 1)

Now for diatomic gas γ = 1.4

Hence C

_{p}= 1004.5 J/Kg*K ≈ 1005 J/Kg*K

6. The isentropic relation between pressure and temperature is given by ____________

a) \((\frac {p_1}{p_2} ) = (\frac {T_1}{T_2} )\)^{1 / (γ – 1)}

b) \((\frac {p_2}{p_1} ) = (\frac {T_2}{T_1} )\)^{1 / (γ – 1)}

c) \((\frac {p_1}{p_2} ) = (\frac {T_1}{T_2} )\)^{γ / (γ – 1)}

d) \((\frac {p_2}{p_1} ) = (\frac {T_2}{T_1} )\)^{γ / (γ – 1)}

View Answer

Explanation: For a perfect gas change in entropy is given by;

ΔS = C

_{p}ln\((\frac {T_2}{T_1} )\) – Rln\((\frac {p_2}{p_1} )\)

However for an isentropic process ΔS = 0.

Therefore C

_{p}ln\((\frac {T_2}{T_1} )\) = Rln\((\frac {p_2}{p_1} )\)

\((\frac {p_2}{p_1} ) = (\frac {T_2}{T_1} )\)

^{Cp/R}

\((\frac {p_2}{p_1} ) = (\frac {T_2}{T_1} )\)

^{γ / (γ – 1)}(As C

_{p}/R = γ / (γ – 1))

7. The following refers to which type of flow?

a) Adiabatic flow

b) Non-adiabatic flow

c) Isentropic flow

d) Reversible flow

View Answer

Explanation: For an adiabatic flow there is no heat transfer during the process. However, since here T

_{0, 2}≠ T

_{0, 1}, the heat transfer occurs during the process, resulting in a change of temperature. Hence the flow is non-adiabatic. Also for a flow to be isentropic, it should be adiabatic, thus it is non-isentropic flow.

8. The air in a reservoir is having a temperature and pressure of 400 K and 36 bar respectively. It expands through duct resulting in a pressure of 8 bar. If the expansion is isentropic, calculate the final temperature.

a) 288.67 K

b) 263.03 K

c) 260.27 K

d) 250.45 K

View Answer

Explanation: For an isentropic process, the relation between pressure and temperature is given by;

\((\frac {p_2}{p_1} )\)

^{(γ – 1)/γ}= \((\frac {T_2}{T_1} )\)

Thus

\((\frac {8}{36})\)

^{(1.4 – 1)/1.4}= \((\frac {T_2}{400} )\)

T

_{2}= 260.27 K

9. The equation of perfect gas is valid for temperatures > 2000 K.

a) True

b) False

View Answer

Explanation: For the equation of perfect gas to be valid, the gas has to be calorically and thermally perfect. However, at temperature > 2000 K, both C

_{p}and C

_{v}become a function of temperatures i.e. ratio of specific heat is no more constant, and gas become calorically and thermally imperfect. Thus for temperature > 2000 K the equation of perfect gas is invalid.

10. When does the process is termed to be an impossible process?

a) ds > 0

b) ds < 0

c) ds = 0

d) ds = Infinite

View Answer

Explanation: According to the second law of thermodynamics, the process always occurs in the direction of increasing entropy. However, for ds < 0, it negates the second law of thermodynamics. Hence for ds < 0, the process is termed as an impossible process. For ds > 0, the process is called irreversible process, and for ds = 0, the process is called a reversible process.

11. The air in a reservoir having a volume of 0.51 m^{3} and temperature 288 K, is compressed to final temperature 350 K. If the change in entropy during the process is -0.3 KJ/K, calculate the final volume of air. Take R = 287 J/Kg*K, and C_{v} = 718 J/Kg*K.

a) 0.83 m^{3}

b) 0.60 m^{3}

c) 0.39 m^{3}

d) 0.11 m^{3}

View Answer

Explanation: For anon-isentropic process;

ΔS = C

_{v}ln\((\frac {T_2}{T_1} )\) + Rln\((\frac {v_2}{v_1} )\)

i.e. -300 = 718 * ln\((\frac {350}{288} )\) + 287 * ln\((\frac {v_2}{0.51} )\)

v

_{2}= 0.11 m

^{3}

12. If heat is added at constant volume __________

a) Enthalpy increases

b) Enthalpy decreases

c) Internal energy increases

d) Internal energy decreases

View Answer

Explanation: The first law of thermodynamics for the constant volume process can be written as;

δq = du = C

_{v}dT

Therefore for a constant volume process the internal energy of the system increase due to heat addition. Whereas for a constant pressure process;

δq = dh = C

_{p}dT Also dh = vdp

Hence for a constant pressure process, both internal energy and enthalpy increases due to heat addition.

13. Calculate the total temperature for an object traveling at M = 0.7, operating at sea level condition. Assume the flow to be isentropic.

a) 295.32 K

b) 316.22 K

c) 350.46 K

d) 396.19 K

View Answer

Explanation: Static temperature at sea level condition (T) = 288 K

Now the ratio of total to static temperature is given by;

\(\frac {T_0}{T}\) = 1 + \((\frac {\gamma – 1}{2})\)M

^{2}

Therefore \(\frac {T_0}{288}\) = 1 + \((\frac {1.4-1}{2})\)0.7

^{2}

T

_{0}= 316.22 K

**Sanfoundry Global Education & Learning Series – Gas Dynamics.**

To practice all areas of Gas Dynamics, __ here is complete set of Multiple Choice Questions and Answers__.