Important question of First Law of Thermodynamics
First Law of Thermodynamics
Quiz
- The first law of thermodynamics was developed by-
- Joule
- Carnot
- Charles
- Kelvin
- Joules law states the specific internal energy of a gas depends only on
- the pressure of the gas
- the volume of the gas
- pressure and volume of the gas
- the temperature of the gas
- Internal energy of an isolated system :
- Increases with addition on heat
- Increases with increase in mass of the system
- Remains same
- Increases with addition of work
- The identity δQ = dU + PdV is valid for :
- Any process occurring in an open system
- Any process occurring in a closed system
- A quasi-static process without dissipation in a
closed cycle - Any process in any system
- Heat transferred to a closed stationary system constant volume is equal to
- increase in enthalpy
- work transfer
- increase in gibbs function
- increased in internal energy
- During throttling process :
- internal energy does not change
- pressure does not change
- volume does not change
- enthalpy does not change
- During an isothermal expansion process of a gas
- pressure remains constant
- temperature remains constant
- both pressure and temperature remain
constant - none of the above
- The change in enthalpy of a closed system is equal to the heat transferred, if the reversible process takes place at constant
- Temperature
- Pressure
- Volume
- Entropy
- Gas expands for a definite volume in a closed vessel. The maximum work will be done when the process is at constant
- Volume
- Temperature
- Pressure
- Enthalpy
- Heat transfer in a cyclic process are +20 kJ, -5 kJ, -10 kJ and +15 kJ. Net work done for this cycle will be given by:
- +0 kJ
- -20 kJ
- +20 kJ
- -10 kJ
- Flow work is analogous to
- Stirring work
- Electrical work
- Displacement work
- Shaft work
- If H be the heat supplied to the system to do work W and change in internal energy as ∆U, then,
- H = ∆U + W
- ∆U = H + W
- W = H + ∆U
- H = W/∆U
- A system undergoes process a process in which the heat transfer to the system is 30 KJ and the work done by the system is 35000 Nm. The change in internal energy of the system is
- +5 KJ
- –10 KJ
- –5 KJ
- + 10 KJ
- Air is compressed isothermally by performing work equal to 16 kJ upon it. The change in internal energy is
- –16 kJ
- Zero
- 16 kJ
- 32 kJ
- A machine produces 100 kJ work by consuming 100 kJ heat. This machine will be called:-
- PMM - I
- PMM - II
- PMM - III
- None of the above
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