An ideal gas is taken from the state A (pressure P, volume V) to the state B (pressure `p//2`, volume 2V) along a straight line path in the P-V diagram. Select the correct statement (s) from the following :

Options

A. The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along the isotherm.
B. In the T-V diagram, the path AB becomes a part of a parabola
C. In the P-T diagram, the path AB becomes a part of a hyperbola
D. In going from A to B, the temperature T of the gas first increases to a maximum value and then decreases.

Answer

A, B, D

Explanation

(a,b,d)
The work done by the gas in the process A to B exceeds the
work that would be done by it if the system were taken from
A to B along the isothermal line. This is because the work
done is the area under the P-V indicator diagram. As shown
by the diagram the area undear the graph in first diagram will
be more than in second diagram. When we extrapolate the
graph shown in figure. (i). Let `P_0` be the intercept on P-axis
and `V_0` be the intercept on V-axis. The equation of the line
AB can be written as
`P=-(P_0)/(V_0)V+P_0......(i) [:' y=mx+c]`


`:. P=-(P_0)/(V_0)xx(RT)/P+P_0 [:' V=(RT)/P]`
`rArrP^2V_0-PP_0V_0=-P_0RT....(ii)`
Relation between P and T is the equation of a parabola.
Also, `P=(RT)/V`
From (i) and (ii)
`(RT)/V=-(P_0)/(V_0)V+P_0 rArr T=-(P_0)/(V_0 R)V^2+(P_0)/R V`
The above equation is of a parabola
(between Tand V)
Differentiating the above equation w.r.t.V, we get
`(dT)/(dV)=-(P_0)/(V_0R)xx2V+(P_0)/R`
For `(dT)/(dV)=0 , V=(V_0)/2`
Also, `(d^2T)/(dV^2)=(-2P_0)/(V_0R)=-ve`
`rArr V=(V_0)/2 is the value for maxima of temperature
Also, `P_AV_A=P_BV_B`
`rArr T_A=T_B` (From Boyle's law)
`rArr` In going from A to B, the temperature of the gas first
increase to a maximum `(at V=(V_0)/2))` and the decreases and
reaches back to the same value.