(a) Its pressure increases (b) Its temperature falls (c) Its density increases (d) Its thermal energy increases Answer-b (5)
Cp/Cv for the gas is (a) 3/2 (b) 4/3 (c) 2 (d) 5/3 Answer-a (10)
(P1,V1,T1) to (P2,V2,T2) . Which of the following relation is correct (a) T1V1γ-1=T2V2γ-1 (b) P1V1γ-1=P2V2γ-1 (c) T1P1γ=T2P2γ (d) T1V1γ=T2V2γ Answer-a (9)
(a) P‘=P (b) P‘=2P (c) P‘=P ×215/2 (d) P‘=7P Answer-c (8)
(a) 27.5°C (b) 75°K (c) 150K (d) 150°C Answer-c (3)
(a) 627°C (b) 527°C (c) 427°C (d) 327°C Answer-d (12)
(a) Adiabatic process (b) Isothermal process (c) Isobaric process (d) Isochoric process Answer-a (4)
(a) 420K (b) 327°C (c) 300K (d) -142°C …
(a) -2767.23 J (b) 2767.23 J (c) 2500J (d) -2500J Answer-b (3)
(a) Isochoric (b) Isobaric (c) Isothermal (d) Adiabatic Answer-(d) In adiabatic process, no transfer of heat takes place between system and surrounding. (4)
(a) Zero (b) Infinite (c) Finite but non-zero (d) Undefined Answer-a (6)
(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)