A mono atomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a friction-less piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1/T2 is given by
(L1 / L2) 2/3
(L1 / L2)
(L2 / L1)
(L2 / L1)2/3
The PT diagram for an ideal gas is shown in the figure, where AC is an adiabatic process. Find the corresponding PV diagram.
One mole of a mono atomic gas is heated at a constant pressure of 1 atmosphere from 0 K to 100 K. If the gas constant R = 8.32J/mol K, the change in internal energy of the gas is approximately.
2.3 J
46 J
8.67 x 103 J
1.25 x 103 J
Which one is correct ?
In an isobaric process, ΔP = 0
In an isochoric process, ΔW = 0
In an isothermal process, ΔT = 0
In an adiabatic process, ΔQ = 0
100 g of water is heated from 30oC to 50oC. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/kg/K)
4.2 kJ
8.4 kJ
84 kJ
Which statement is incorrect ?
all reversible cycles have same efficiency
reversible cycle has more efficiency than an irreversible one
Carnot cycle is a reversible one
Carnot cycle has the maximum efficiency of all the cycles
A thermodynamic process is shown in the figure. The pressure and volumes corresponding to some points in the figure are
PA = 3 × 104 Pa, VA = 2 × 10-3 m3
PB = 8 × 104 Pa, VB = 5 × 10-3 m3
In process AB, 600 J of heat is added to the system and in process BC, 200 J of heat is added to the system. The change in internal energy of the system in process AC would be
560 J
800 J
600 J
640 J
In an adiabatic system which is true ?
PγTγ-1 = constant
PγT1-γ = constant
PTγ = constant
P1-γTγ = constant
Even Carnot engine cannot give 100% efficiency, because we cannot
prevent radiation
find ideal sources
reach absolute zero temperature
eliminate friction
When a system is taken from the initial state i to final state f along he path iaf, it is found that Q = 50 cal and W = 20 cal. If along the path ibf, Q = 36 cal, then W along the path ibf is
6 cal
16 cal
66 cal
14 cal
