Test: Basic Concept - 3 - Mechanical Engineering MCQ

The correct answer is 3: 1, 2, and 4.

1. The study of thermodynamics can determine whether a process is feasible or not. This is done by analyzing the energy changes and entropy changes associated with the process. If the energy change is positive (indicating energy input) and the entropy change is negative (indicating a decrease in disorder), the process may not be feasible.

2. Thermodynamics allows us to quantify the energy required for a process. By analyzing the energy changes (such as heat and work) associated with a reaction or process, we can determine the amount of energy needed for it to occur.

3. The study of thermodynamics does not directly provide information about the rate or speed with which a process occurs. However, it can provide insights into the factors that influence reaction rates, such as temperature and pressure. This is because the rate of a reaction is influenced by the energy changes involved (activation energy) and the thermodynamic properties of the reactants and products.

4. Thermodynamics also provides information about the extent to which a reaction or process takes place. This is typically determined by the concept of equilibrium, which is governed by the thermodynamic properties of the system. The equilibrium constant of a reaction is related to the extent to which reactants are converted into products at equilibrium.

We know that for an Ideal gas

PV = mRT

Also P_abs = P_atm + P_g = 1 + 0.5        

= 1.5 bar = 150 kPa

Volume = 2.87 m3

R = 0.287 kPa

T = 300 K
Substituting the values, we get 1

150 × 2.87 = m × 0.287 × 300 m

= 5.00 kg

Hence, the correct option is (c).

Let us draw process in p-v diagram

Process 1-2: Constant pressure

Process 2-3: Constant volume

Hence, the correct option is (b).

In a quasiequilibrium process, the pressure in a system is constant at any given instant throughout the system. This is because a quasiequilibrium process is a thermodynamic process that occurs slowly enough for the system to remain in internal equilibrium. As the process progresses, the pressure across the system may change, but at any instant, it remains constant. 

Given: C₁ = 10 m/s, C₂ = 20 m/s, z₂ – z₁ = 20 m,
T₂ – T₁ = 1 K
C_V = 100 J/kg-K, g = 10 m/s², δQ = 0

Assuming compressible fluid with γ = 1

SFEE:

On the system, δ W = 4.5 kJ

Hence, the correct answer is 4.5.

50 cm Hg vacuume:

80 cm Hg gauge:

1.2 m of H20 guage:

Properties are the coordinates to describe the state of a system. They are state variables of the system. Any operation in which one or more of the properties of a system changes is called a change of state. The succession of states passes through during a change of state is called the path of the change of state is called a process. A thermodynamic cycle is defined as a series of state changes such that the final state is identical with the initial state.

Net work done = 

Work interaction or work done at boundary of system is only by drag force

W = −F_d × s = −0.5 × 50 = −25 J

This is work done by surrounding air on system and is negative. Hence, positive work is done by body on surrounding air and is equal to 25 J.
Hence, the correct option is (d).

p₂ = p_gauge + p_atom = 1 + 1.013

= 2.013 bar

As volume is constant, we use ideal gas state

Initial p_gauge = p₁ − p_atm = 2.0854 − 1.0132     

= 1.07 bar

Hence, the correct option is (c).