(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 …
(a) Internal energy of the gas increases (b) Internal energy of the gas decreases (c) Internal energy remains unchanged (d) Average kinetic energy of gas molecule decreases Ans. c (12)
(a) Never (b) Yes (c) They will cut when temperature is 0°C (d) Yes, when the pressure is critical pressure Ans. a (8)
(a) Heat content remains constant (b) Heat content and temperature remain constant (c) Temperature remains constant (d) None of the above Ans. c (11)
(a) It introduces the concept of the internal energy (b) It introduces the concept of the entropy (c) It is not applicable to any cyclic process (d) None of the above Ans. b (20)
(a) Zero (b) Negligible (c) Negative (d) Positive Ans. c (5)
8 x 105 J of heat and doing 6.5 x 105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105J of heat. Then in the second process (a) Work done …
(a) Pressure only (b) Volume only (c) Pressure, volume and temperature (d) Number of moles Ans. c (10)
(a) More for (i) (b) More for (ii) (c) Same in both cases (d) Independent of number of moles Ans. c (6)
(a) 18J (b) 9J (c) 4.5J (d) 36J Ans .a (5)
(a) 19.1 kJ (b) 12.5 kJ (c) 25 kJ (d) Zero Ans. a (6)
(a) 19.1 kJ (b) 12.5 kJ (c) 25 kJ (d) Zero Ans. a (6)
at constant pressure 2.1 x 105N/m2 , there was an increase in its volume equal to 2.5 x 10-3 m3. The increase in internal energy of the gas in Joules is (a) 450 …
(a) Zeroth law of thermodynamics (b) First law of thermodynamics (c) Second law of thermodynamics (d) Newton’s law of cooling Ans. a (5)
to that at constant volume is γ, the change in internal energy of a mass of gas, when the volume changes from V to 2V constant pressure p, is (a) R/( γ – 1) …
(a) Its temperature will increase (b) Its temperature will decrease (c) Its temperature will remain constant (d) None of these Ans. a (5)
(a) Increases by 600 J (b) Decreases by 800 J (c) Increases by 800 J (d) Decreases by 50 J Ans. c (7)
(a) 7 : 6 (b) 6 : 7 (c) 36 : 49 (d) 49 : 36 Answer-b (159)