(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)
1×105 N/m2 pressure is converted into ice of volume 1.091 cm3 , the external work done will be (a) 0.0091 joule (b) 0.0182 joule (c) – 0.0091 joule (d) – 0.0182 joule Answer-a (5)
(a) External work done (b) Rise in temperature (c) Increase in internal energy (d) External work done and also rise in temp. Answer-a (5)
(a) No work is done against gas (b) Heat is relased by the gas (c) The internal energy of gas will increase (d) Pressure does not change Explanation-(b) In isothermal process, heat is released by the gas to maintain the …
(a) Isochoric process (b) Isothermal process (c) Isobaric process (d) None of these Explanation-(b) In isothermal process, temperature remains constant. (6)
(a) Internal energy of the system decreases (b) Work done by the gas is positive (c) Work done by the gas is negative (d) Internal energy of the system increases Answer-c (5)
(a) Copper (b) Glass (c) Wood (d) Cloth Explanation-(a) An isothermal process takes place at constant temperature, must be carried out in a vessel with conducting wall so that heat generated should go out at once. (7)
(a) Decrease (b) Increase (c) Remain same (d) None of the above Answer-c (10)
(a) Infinite (b) Zero (c) Negative (d) Remains constant Answer-a (12)
(a) Temperature is constant (b) Internal energy is constant (c) No exchange of energy (d) (a) and (b) are correct Explanation-(c) In isothermal process, exchange of energy takes place between system and surrounding to maintain the system temperature constant. (13)
(a) Boyle’s law (b) Charle’s law (c) Gaylussac law (d) None of the above Answer-a (13)
(a) 1.013×105 N/m2 (b) 1.013×106 N/m2 (c) 1.013×10-11 N/m2 (d) 1.013×1011 N/m2 Answer-a (9)
(a) RT log10 (V2/V1) (b) RT log10 (V1/V2) (c) RT loge (V2/V1) (d) RT loge (V1/V2) Answer-c (12)
(a) E/4 (b) E/16 (c) E (d) 4 E Answer-c (8)
(a) Temperature only (b) Compressibility only (c) Both temperature and compressibility (d) None of these Answer-b (7)
(a) P (b) γ (c) P / 2 (d) P /γ Answer-a (9)
(a) Remains constant (b) Increases linearly with temperature (c) Decreases linearly with temperature (d) Decreases inversely with temperature Answer-a (12)
O2 gas having a volume equal to 22.4 litres at 0°C and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is (a) 1672.5 J …
(a) Isothermal changes only (b) Adiabatic changes only (c) Both isothermal and adiabatic changes (d) Neither isothermal nor adiabatic changes Answer-c (8)
the value of ΔP/P is equal (a) – γ1/2 ΔV/V (b) -(ΔV/V) (c) -γ1 ΔV/V (d) -γ2 ΔV/V Ans. b (7)
(a) 4/3 m (b) 0.5 m (c) 2.0 m (d) 3/4 m Ans. b (8)
of 96 gm of oxygen at 27°C is increased from 70 litres to 140 litres, then the work done by the gas will be (a) 300 R log102 (b) 81 R loge2 …