(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)
(a) Kinetic energy only (b) Kinetic and potential energy (c) Potential energy (d) None of these Ans. a (4)
(a) P (b) T (c) V (d) R Ans. d (9)
(a) Enthalpy (b) Work done (c) Gibb’s energy (d) Internal energy Ans. b (12)
(a) 150 J (b) 70 J (c) 110 J (d) 40 J Ans. b (5)
1m3 is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at 300 K. The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and …
Δw when an amount of heat is ΔQ added to the system, the corresponding change in the internal energy is ΔU . A unique function of the initial and final states (irrespective of the mode of change) is (a) …
(a) 7900 J (b) 8200 J (c) 5600 J (d) 6400 J Ans. a (5)
of one atmosphere from 00c to 1000c . Then the change in the internal energy is (a) 6.56 joules (b) 8.32 x 102 joules (c) 12.48 x 102 joules (d) 20.80 joules Ans. c (15)
(a) 0.56×104 J (b) 1.3×102 J (c) 2.7×102 J (d) 3.4×103 J Answer-d (9)
(a) 4.6×103 K (b) 11.6×103 K (c) 23.2×103 K (d) 7.7×103 K Answer-d (4)
(a) Newton’s law (b) Law of conservation of energy (c) Charle’s law (d) Law of heat exchange Ans. b (5)
(a) (3/2)kT (b) kT (c) (1/2)kT (d) (3/2)RT Answer-c …
(a) Only potential energy (b) Only kinetic energy (c) Potential and kinetic energy both (d) None of the above Answer-b (10)
(a) Temperature (b) Energy (c) Work (d) None of these Ans. a (9)
T1 and T2 are mixed. There is no loss of energy. The masses of the molecules are m1 and m2 and the number of molecules in the gases are n1 and n2 The temperature of mixture will be (a) (T1 …
(a) 2×√3R×300 gm×cm/sec (b) 2×3×R×300 gm×cm/sec (c) 1×√3R×300 gm×cm/sec (d) Zero Answer-d (10)
(a) 30 J (b) 20 J (c) 60 J (d) 40 J Ans. c (5)
(a) P (b) < P (c) > P (d) Zero Answer-c (4)
(a) Increase continuously (b) Decreases continuously (c) First increases and then decreases (d) First increase and then becomes constant Answer-c (4)
CO2 mixture at 27°C. If R = 8.31 J/mole kelvin, then the pressure in the vessel in N/m2 will be (approx.) (a) 5×105 (b) 5×104 (c) 106 …
(a) The internal energy changes in all processes (b) Internal energy and entropy are state functions (c) The change in entropy can never be zero (d) The work done in an adiabatic process is always zero (19)
(a) 100 J (b) 3000 J (c) 450 J (d) 150 J Answer-d (17)
(a) Transfer momentum to the walls (b) Momentum becomes zero (c) Move in opposite directions (d) Perform Brownian motion Answer-a (24)
(a) Volume (b) Temperature (c) Pressure (d) Work Ans. d (4)
is Cp=3.4 x 103 cal/kg 0c and at constant volume is Cv= 2.4 x103 cal/kg 0c. If one kilogram hydrogen gas is heated from 10 0c to 200c at constant pressure, the external work done on the gas to maintain …
(a) The temperature will decrease (b) The volume will increase (c) The pressure will remain constant (d) The temperature will increase Ans. a (7)
(a) Pressure and volume (b) Volume and temperature (c) Temperature and pressure (d) Any one of pressure, volume or temperature Ans. d (12)
(a) Q=W=0 and ΔEint =0 (b) Q=0,W>0 and ΔEint =-W (c) W=0,Q>0, and ΔEint = Q (d) W>0,Q<0 and ΔEint =0 Ans. a (8)
(a) ΔQ= ΔU + ΔW (b) ΔQ= ΔU – ΔW (c) ΔQ= ΔW-ΔU (d) ΔQ= – ΔW-ΔU Ans. b (6)
(a) 260 J (b) 150 J (c) 110 J (d) 40 J Ans. d (8)