The correct option is
C Internal energy of the gas at state
a is greater than that at state
c From the ideal gas equation,
PM=ρRT
⇒Pρ=RTM
Here,
RTM is the slope of
P−ρ curve.
As
R and
M are constant for a gas,
Slope
(m)∝T
From graph,
mcd>mab, thus,
Tc>Ta
Also, the slope of the processes
ab and
cd are constant, thus, temperature is constant in these processes.
⇒ ab and
cd are isothermal processes.
For a given gas,
ρ=MV.
For processes
bc and
da, density is constant, thus, volume will be constant.
⇒ bc and
da are isochoric processes.
Plotting the
P−V diagram, we get,
For option
(a),
In process
bc, as there is no change in volume, the work done by gas is zero.
For option
(b),
In process
cd, as volume is decreasing and pressure is increasing, thus it is a compression process. Therefore work done by gas is negative.
For option
(c),
As
Tc>Ta, internal energy at
c is greater than at
a.
For option
(d),
As the process is cyclic anticlockwise, thus, the work done by gas is negative.
Hence, option
(c) contains incorrect statement.
Why this question? It provides unique challenge to apply laws of thermodynamics and ideal gas equation. Caution: For comparing work done and change in internal energy in a given process, always convert given curve to P - V diagram.