(a) The work is done by the gas (b) Internal energy of the gas increases (c) Both (a) and (b) (d) None from (a) and (b) Answer-c (6)
V1 to V2. The work done by the gas is (a) P(V2-V1) (b) P(V1-V2) (c) P(V1γ-V2γ) (d) (PV1V2)/(V2-V1) Answer-a (6)
2.1×105N/m2. What will be its isothermal modulus of elasticity[(Cp/Cv)=1.4] (a) 1.8×105N/m2 (b) 1.5×105N/m2 (c) 1.4×105N/m2 (d) 1.2×105N/m2 Answer-b (14)
(a) Zero (b) 100J (c) – 50J (d) 50J Answer-d (8)
(a) 350×40.4 K (b) 300×40.4 K (c) 150×40.4 K (d) None of these Answer-b (2)
(a) 1 : 16 (b) 1 : 8 (c) 1 : 4 (d) 8 : 1 Answer-d (3)
(a) Lesss than P (b) More than P (c) P (d) Either (a) or (c) Answer-b (6)
(a) ΔU=-W in an adiabatic process (b) ΔU=W in an isothermal process (c) ΔU=-W in an isothermal process (d) ΔU=W in an adiabatic process Answer-a (2)
(a) Adiabatic expansion (b) Adiabatic compression (c) Isothermal expansion (d) Isothermal compression Answer-b (5)
105 N/m2 . On suddenly pressing the piston the volume is reduced to half the initial volume. The final pressure of the gas is (a) 20.7 ×105 (b) 21.3 ×105 …
(a) 273 K (b) 573 K (c) 373 K (d) 473 K Answer-a (6)
(a) (T+2.4)K (b) (T-2.4)K (c) (T+4)K (d) (T-4)K Answer-d (8)
(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 (11)
(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 (10)
(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 (9)
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 (63)
(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 (7)
(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 (11)
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)