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 = Cp 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 Cp and Cv are constant and are independent of temperature. Now for a calorically perfect gas, both Cp and Cv are constant and for a thermally perfect gas Cp and Cv 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 = Cp – Cv
R / Cp = 1 – Cv / Cp
R / Cp = (γ – 1) / γ (Since, γ= Cp / Cv)
Cp = γR / (γ – 1)
Now for diatomic gas γ = 1.4
Hence Cp = 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 = Cpln\((\frac {T_2}{T_1} )\) – Rln\((\frac {p_2}{p_1} )\)
However for an isentropic process ΔS = 0.
Therefore Cpln\((\frac {T_2}{T_1} )\) = 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 Cp/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 T0, 2 ≠ T0, 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 = \((\frac {T_2}{400} )\)
T2 = 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 Cp and Cv 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 m3 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 Cv = 718 J/Kg*K.
a) 0.83 m3
b) 0.60 m3
c) 0.39 m3
d) 0.11 m3
View Answer
Explanation: For anon-isentropic process;
ΔS = Cvln\((\frac {T_2}{T_1} )\) + Rln\((\frac {v_2}{v_1} )\)
i.e. -300 = 718 * ln\((\frac {350}{288} )\) + 287 * ln\((\frac {v_2}{0.51} )\)
v2 = 0.11 m3
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 = CvdT
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 = CpdT 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 + \((\frac {\gamma – 1}{2})\)M2
Therefore \(\frac {T_0}{288}\) = 1 + \((\frac {1.4-1}{2})\)0.72
T0 = 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.
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