# Rocket Propulsion Questions and Answers – Chemical Rocket Propellant – Background and Fundamentals

This set of Rocket Propulsion written test Questions & Answers focuses on “Chemical Rocket Propellant – Background and Fundamentals”.

1. Consider a mixture of three gases a, b and c at equilibrium. If the individual gas components have pressures equal to Pa, Pb and Pc, determine the total pressure P of the mixture of gases.
a) P = Pa + Pb + Pc
b) P = Pa x Pb x Pc /(Pa + Pb + Pc)
c) P = [(Pa x Pb) + (Pb x Pc) + (Pc x Pa)] / (Pa + Pb + Pc)
d) P = (Pa x Pb / Pc) + (Pb x Pc / Pa) + (Pc x Pa / Pb)
View Answer

Answer: a
Explanation: Dalton’s law of partial pressures states that the total pressure exerted by a mixture of gases is the sum of partial pressures of individual gas components. Hence the answer is P = Pa + Pb + Pc.
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2. Perfect gas equation applies very closely to ________ gases.
a) low temperature, low pressure
b) high temperature, high pressure
c) high temperature, low pressure
d) low temperature, high pressure
View Answer

Answer: c
Explanation: Perfect gas equations are valid for high temperature and low-pressure gases. In such a case, the inter-molecular forces are negligibly small. PV=RT is the perfect gas equation, with V as the specific volume, P as the pressure, T as the temperature and R as the gas constant for the mixture of gases.

3. Consider two sets of mixture of gases A and B having the same specific volume of the mixture and having total pressures PA and PB respectively. If TA = 3TB and molecular mass of A MA = 1/2 MB, determine the ratio of PA/PB.
a) 0.75
b) 12
c) 1.5
d) 6
View Answer

Answer: d
Explanation: Assume that the gases follow the perfect gas equation. P = R’T/MVmix where Vmix is the specific volume of the mixture and R’ is the universal gas constant.
So PA/PB = (TA/TB) x (MB/MA)
= 3 x 2 = 6.

4. Consider two sets of mixture of gases X and Y having the same total temperature of the mixture and having total pressures PX and PY respectively. If PX = 1.4 PY and molecular mass of A MX = 2.3 x MY, determine the ratio of Vmix_X/Vmix_Y.
a) 0.26
b) 0.52
c) 0.62
d) 0.31
View Answer

Answer: d
Explanation: Assuming the mixture of gases to be ideal, we have P = R’T/MVmix where Vmix is the specific volume of the mixture and R’ is the universal gas constant.
PX/PY = (TX/TY) x (MYVmix_Y/MXVmix_X)
Vmix_X/Vmix_Y = (MY/MX) x (TX/TY) x (PY/PX)
= (1/2.3) x 1 x (1/1.4) = 0.31.

5. In a gaseous mixture, the number of moles and the molecular mass (in their respective units) of each component is given as follows: nA = 1, MA = 23; nB = 2, MB = 48; nC = 3, MC = 19. Find the average molecular mass of the mixture.
a) 28.67
b) 29.33
c) 31.25
d) 25.47
View Answer

Answer: b
Explanation: Mavg = (nAMA + nBMB + nCMC) / (nA + nB + nC)
= (23 + 96 + 57) / 6 = 29.33.
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6. What is the number of condensed species in a gaseous mixture, if there are 8 possible species which enter into a relationship and out of these, only 5 are gases?
a) 3
b) 2
c) 9
d) 6
View Answer

Answer: a
Explanation: If there are n species that enter into a relationship and if there are only m species out of these, then the total number of condensed species is n-m.

7. Determine the stoichiometric mixture mass ratio for the complete reaction: H2 + 1/2 O2 → H2O.
a) 2:1
b) 16:1
c) 4:1
d) 8:1
View Answer

Answer: d
Explanation: On mass basis, stoichiometric mixture requires half of 32 kg of O2 with 2 kg of H2. So the mixture mass ratio would be 1/2 x 32 / 2 = 8.

8. Rocket propulsion usually operates in ___________ mixture ratio.
a) fuel rich
b) oxidizer rich
c) stoichiometric
d) any
View Answer

Answer: a
Explanation: Rocket propulsion systems typically operate in fuel rich mixture ratio. This allows a portion of lightweight molecules such as hydrogen to be unreacted and contributes to the decrease in the average molecular mass of the combustion products.

9. The decomposition of solid propellants to reaction mixture is a(n) ______________ chemical reaction.
a) reversible
b) irreversible
c) stoichiometric
d) ideal
View Answer

Answer: b
Explanation: The whole process of combustion of solid propellants to derive a reaction mixture is an irreversible chemical reaction. It means that the reverse process is not possible.
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10. What is the energy released (or absorbed), or the enthalpy change when one mole of a chemical compound is formed from its constituent atoms or elements at one bar and isothermally at 25°C?
a) Heat of formation
b) Heat of reaction
c) Gibbs free energy
d) Latent heat of vaporization
View Answer

Answer: a
Explanation: This is the definition of heat of formation. In this reaction, each of the reactants and products are in its thermodynamic standard state and at the reference pressure and temperature.

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advertisement Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Instagram | Facebook | Twitter