The equation of state for 5g of oxygen at a pressure p and temperature T, when occupying a volume V, will be
pV = (5/32)RT
pV = 5 RT
pV = (5/2)RT
pV = (5/16)RT
The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
1/3
1/2
2/3
1/4
A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is
20 J
-20 J
400 J
-374 J
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
An ideal mono atomic gas is taken around the cycle ABCDA as shown in the PV diagram. The work done during the cycle is given by
1/2 PV
PV
2 PV
4 PV
If cp and cv denote the specific heat of nitrogen per unit mass at constant pressure and constant volume respectively, then
cp - cv = R/28
cp - cv = R/14
cp - cv = R
cp - cv = 28R
An ideal gas undergoing an adiabatic change has the following pressure-temperature relationship
pγ-1Tγ = constant
pγTγ-1 = constant
pγT1-γ = constant
p1-γTγ = constant
The volume (V) versus temperature (T) graphs for a certain amount of a perfect gas at two pressures P1 and P2are shown in the figure. It follows from the graph that
P1 > P2
P1 < P2
P1 = P2
information is insufficient to draw any conclusion.
A gas at state A changes to state B through path I and II shown in figure. The changes in internal energy are ΔU1 and ΔU2 respectively. Then
Δ U1 > ΔU2
Δ U1 < ΔU2
Δ U1 = ΔU2
Δ U1 = ΔU2 = 0
First law of thermodynamics is a consequence of conservation of
Work
Energy
Heat
All of these
