(a) 10°C (b) 887°C (c) 668K (d) 144°C …
(a) dQ=0 (b) dU=-dW (c) Q = constant (d) Entropy is not constant Answer-d (9)
(a) 1280 J (b) 1610 J (c) 1815 J (d) 2025 J Answer-c (6)
T to T1 is (a) R(T-T1) (b) [R/(γ-1)](T-T1) (c) RT (d) R(γ-1)(T-T1) Answer-b (8)
change is ( γ=Cp/Cv ) (a) PTγ constant (b) PT-1+γ constant (c) Pγ-1 Tγ constant (d)P1-γ Tγ constant Answer-d (57)
(a) Isothermal curve slope = adiabatic curve slope (b) Isothermal curve slope =γ× adiabatic curve slope (c) Adiabatic curve slope = γ× isothermal curve slope (d) Adiabatic curve slope =(1/2)× isothermal curve slope Answer-c (6)
(Pi,Vi,Ti) to its final state (Pf,Vf,Tf) . The decrease in the internal energy associated with this expansion is equal to (a) Cv(Ti-Tf) (b) CP(Ti-Tf) (c) (1/2)(Cp+Cv)(Ti-Tf) …
(a) Isothermal process (b) Isobaric process (c) Isochoric process (d) Adiabatic process Ans. d (10)
270 has a volume of 8 litres.It is suddenly compressed to a volume of 1 litre. The temperature of the gas will be [ϒ = 5/3] (a) 1080C …
are T1 and T2 respectively, then the change in internal energy of the gas is (a) (R/ϒ-1) (T2-T1) (b) (R/ϒ-1) (T1-T2) (c) R(T1-T2) …
(a) Zero (b) –100 J (c) 200 J (d) 100 J Ans. d (14)
(a) Isothermal change (b) Adiabatic change (c) Isobaric change (d) Isochoric change Answer-b (7)
(a) 227.36 K (b) 500.30 K (c) 454.76 K (d) -47°C Answer-a (5)
(γ=Cp/Cv) (a) PγV = constant (b)TγV = constant (c) TVγ-1 =constant (d) TVγ = constant Answer-c (6)
P0 , its new pressure will be (a) 8P0 (b) 16P0 (c) 6P0 (d) 2P0 Answer-b (2)
(a) P (b) 2P (c) P/2 (d) γP Answer-d (8)
(a) Starts becoming hotter (b) Remains at the same temperature (c) Starts becoming cooler (d) May become hotter or cooler depending upon the amount of water vapour present Answer-c (3)
(a) 2077.5 joules (b) 207.5 joules (c) 207.5 ergs (d) None of the above Answer-a (18)
(a) No energy is required for expansion (b) Energy is required and it comes from the wall of the container of the gas (c) Internal energy of the gas is used in doing work (d) Law of conservation of energy …
(a) The pressure decreases (b) The pressure increases (c) The pressure remains the same (d) The pressure may increase or decrease depending upon the nature of the gas Answer-a (5)
(a) 1×105 N/m2 (b) 1×10-8 N/m2 (c) 1.4 N/m2 (d) 1.4×105 N/m2 Answer-d (8)
(a) 1 J (b) -1 J (c) 2 J (d) – 2 J Answer-d (4)
(a) 24/5 (b) 8 (c) 40/3 (d) 32 times its initial pressure Answer-d (6)
1.5×104 J . During this process about (a) 3.6×103 cal of heat flowed out from the gas (b) 3.6×103 cal of heat flowed into the gas (c) 1.5×104 cal of heat flowed into the gas (d) 1.5×104 cal of heat …
(a) For an isothermal change PV = constant (b) In an isothermal process the change in internal energy must be equal to the work done (c) For an adiabatic change P2/P1=(V2/V1)γ , where γ is the ratio of specific heats (d) …
(a) 450 K (b) 375 K (c) 225 K (d) 405 K Answer-b (4)
(a) More in the isothermal process (b) More in the adiabatic process (c) Neither of them (d) Equal in both processes Answer-a (8)
(a) 1/128 (b) 32 (c) 128 (d) None of the above Answer-c (7)
If γ is supposed to be (3/2) , then the final pressure is (a) 4 atmosphere (b) 3/2 atmosphere (c) 8 atmosphere (d) 1/4 atmosphere Ans. c (7)
at 300 K. If this tyre suddenly bursts, its new temperature will be ( γ=1.4) (a) 300 (4)1.4/0.4 (b) 300(1/4)-0.4/1.4 (c) 300(2)-0.4/1.4 (d) 300(4)-0.4/1.4 …
(a) ΔU = 0 (b) ΔU = negative (c) ΔU= positive (d) ΔW = zero Ans. b (8)
(a) Change is pressure (b) Change is volume (c) Change in temperature (d) None of the above Ans. a (7)
(a) Reversible adiabatic change and fall of temperature (b) Reversible adiabatic change and rise of temperature (c) Irreversible adiabatic change and fall of temperature (d) Irreversible adiabatic change and rise of temperature Answer-a (9)
(a) Its internal energy decreases (b) Its internal energy does not change (c) The work done by the gas is equal to the quantity of heat absorbed by it (d) Both (b) and (c) are correct Ans. d (14)
The volume of gas is made 900 cm3 by compressing it isothermally. The stress of the gas will be (a) 8 cm (mercury) (b) 7 cm (mercury) (c) 6 cm (mercury) (d) 4 cm (mercury) Ans. a (5)
27°C. The piston is pushed so slowly that the air within the cylinder remains in thermal equilibrium with the surroundings. Find the approximate work done by the system if the final volume is twice the initial volume (a) 543 J …
(R = 8.31 J/mole–K) (a) 750 joules (b) 1728 joules (c) 1500 joules (d) 3456 joules Answer-b (12)
(a) 540 cal (b) 40 cal (c) Zero cal (d) 500 cal Answer-b (5)
(a) 2408 J (b) 2240 J (c) 2072 J (d) 1904 J Answer-c (7)