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1.0.0
MCQ
21. Change in entropy in a irreversible process-
Ds>d𝘘
ds< d𝘘/T
Ds= d𝘘/T
De ɑ d𝘘 /T
22. Which is true for adiabatic process?
∆W = 0
∆Q = 0
∆W=0
∆P=0
23. The basis for measuring thermodynamic property of temperature is given by-
Zeroth law of thermodynamics
First law of thermodynamics
Second law of thermodynamics
Third law of thermodynamics
24. Heat and work are-
Point functions
System properties
Path functions
Intensive properties
25. Calorie is a measure of-
Specific heat
Quantity of heat
Thermal capacity
Entropy
26. Workdone in free expansion process is-
Zero
Minimum
Maximum
Positive
There is no restraining/opposing force or pressure as expansion occurs against vacuum. So, dw = 0 (in case of free expansion).
27. A diathermic wall is one which-
Prevents thermal interaction
Permits thermal interaction
Encourages thermal interaction
Discourages thermal interaction
28. Which of the following is the extensive property of a themodynamic system?
Pressure
Volume
Temperature
Density
29. Which of the following is not the intensive property?
Pressure
Temperature
Density
Heat
30. An isobaric process, has constant-
Density
Temperature
Pressure
Volume
31. Addition of heat at constant pressure to a gas results in-
Raising its temperature
Raising its pressure
Raising its volume
Raising its temperature and doing external work
32. An ideal gas obeys the law-
PT = nR ̅V
PV=n/R ̅T
PV= nR ̅T
PV= nR ̅/T
33. The entropy in an irreversible cyclic process.
Remain constant
Decreases
Increase
First increase then decrease
34. A path 1-2-3 is given. A system absorbs 100 kj as heat and does 60kj of work while along the path 1-4-3, it does 20kj of work. The heat absorbed during the cycle 1-4-3 is-
-140kj
-80kj
-40kj
+60kj
For 1-2-3 cycle. 𝘘1-2-3 = 100 kj W1-2-3 = 60 kj We know, Heat = Internal energy + Work 𝘘1-2-3 = 𝘘1-2-3 + W1-2-3 𝘘1-2-3 = 𝘘1-2-3 - W1-2-3 𝘘1-2-3 = (100-60) kj = 40 kj Again, For 1-4-3 cycle, W1-4-3 = 20kj 𝘘1-2-3 = 𝘘1-4-3 =40kj Again, we know, 𝘘1-4-3 = 𝘘1-4-3 + W1-4-3 = 40 kj+20 kj = 60 kj For 1-2-3 cycle. 𝘘1-2-3 = 100 kj W1-2-3 = 60 kj We know, Heat = Internal energy + Work 𝘘1-2-3 = 𝘘1-2-3 + W1-2-3 𝘘1-2-3 = 𝘘1-2-3 - W1-2-3 𝘘1-2-3 = (100-60) kj = 40 kj Again, For 1-4-3 cycle, W1-4-3 = 20kj 𝘘1-2-3 = 𝘘1-4-3 =40kj Again, we know, 𝘘1-4-3 = 𝘘1-4-3 + W1-4-3 = 40 kj+20 kj = 60 kj
35. The term N.T.P. stands for-
Nominal temperature and pressure
Natural temperature and pressure
Normal temperature and pressure
Normal thermodynamic practice
36. An adiabatic wall is one which-
Prevents thermal interaction
Permits thermal interaction
Encourages thermal interaction
Discourages thermal interaction
37. If the value of n = 0 in the equation PV^n = C, then the process is called-
Constant volume process
Adiabatic process
Constant pressure process
Isothermal process
Constant Pressure Process (n= 0) Isothermal Process (n = 1) [BREB-2014] Adiabatic Process (n = y) Polytropic Process (y> n > 1) Constant Volume Process (n = ∞) PV°=C ⇒ P=C [V°=1]
38. Total heat of a substance is also known as-
Internal energy
Entropy
Thermal capacity
Enthalpy
39. If heat be exchanged in a reversible manner, which of the following property of the working substance will change accordingly?
Temperature
Enthalpy
Internal energy
Entropy
40. In a throttling process-
In a throttling process-
Temperature remains constant
Enthalpy remains constant
Volume remains constant