A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is
(a) ρoVo
(b) 2ρoVo
(c) ρoVo/2
(d)zero
Concept Questions :-
Work done in the cyclic process = Area bound by the closed configuration =Area of closed configuration
={2 [1/2 (Vo/2) x ρo]} + {-2[1/2(Vo/2)ρo]}
=zero
Difficulty Level:
If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cyclic process, then
(a) W=0 (b) Q=W=0
(c) E=0 (d) Q=0
Concept Questions :-
In a cyclic process, system is not isolated from the surroundings.
In a cyclic process, a system starts from one point and ends at the same point. In this case, the change in the internal energy must again be zero and therefore the thermal energy added to the system must equal the work done during the cycle. That is, in a cyclic process.
Difficulty Level:
In a cyclic process, the internal energy of the gas -
(1) Increases
(2) Decreases
(3) Remains constant
(4) Becomes zero
Concept Questions :-
(3) Internal energy is a state function.
Difficulty Level:
A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is -
(1) 20 J
(2) – 20 J
(3) 400 J
(4) – 374 J
Concept Questions :-
(2) Work done by the system = Area of shaded portion on P-V diagram
Since sense is anti-clockwise, work done is negative
Difficulty Level:
An ideal gas is taken around ABCA as shown in the above P-V diagram. The work done during a cycle is
(1) 2PV
(2) PV
(3) 1/2PV
(4) Zero
Concept Questions :-
(1) Work done = Area enclosed by triangle ABC
Difficulty Level:
An ideal gas is taken through the cycle A → B → C → A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C → A is
(1) – 5 J
(2) – 10 J
(3) – 15 J
(4) – 20 J
Concept Questions :-
(1) For cyclic process. Total work done
ΔWAB = PΔV = 10(2 – 1) = 10J and ΔWBC =0
(as V = constant)
ΔQ = ΔU + ΔW
ΔU = 0 (Process ABCA is cyclic)
⇒ ΔQ = ΔWAB + ΔWBC + ΔWCA
⇒ 5 = 10 + 0 + ΔWCA ⇒ ΔWCA = – 5 J
Difficulty Level:
An ideal gas is taken around the cycle ABCA as shown in the P-V diagram. The net work done by the gas during the cycle is equal to
(1) 12 P1V1
(2) 6 P1V1
(3) 3 P1V1
(4) 2 P1V1
Concept Questions :-
(4) Work done
Difficulty Level:
Heat energy absorbed by a system in going through a cyclic process shown in figure is
(1) 107 π J
(2) 104 π J
(3) 102π J
(4) 10–3π J
Concept Questions :-
(3) In a cyclic, ΔU = 0
Δ Q = ΔU + ΔW = 0 + ΔW = Area of closed curve
Difficulty Level:
In the diagrams (i) to (iv) , variation of volume with changing pressure is shown. A gas is taken along the path ABCD. The change in internal energy of the gas will be
(1) Positive in all cases (i) to (iv)
(2) Positive in cases (i), (ii) and (iii) but zero in (iv) case
(3) Negative in cases (i), (ii) and (iii) but zero in (iv) case
(4) Zero in all four cases
Concept Questions :-
(4) In all given cases, process is cyclic and in cyclic process ΔU = 0.
Difficulty Level:
Carbon monoxide is carried around a closed cycle abc in which bc is an isothermal process as shown in the figure. The gas absorbs 7000 J of heat as its temperature increases from 300 K to 1000 K in going from a to b. The quantity of heat rejected by the gas during the process ca is -
(1) 4200 J
(2) 5000 J
(3) 9000 J
(4) 9800 J
Concept Questions :-
(4) For path ab :
By using
For path ca :
….(i)
∵
∴ ….(ii)
Also
….(iii)
on solving equations (i), (ii) and (iii)
= –9800 J
Difficulty Level: