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Question

One mole of gas expands obeying the relation as shown in the P/V diagram. The maximum temperature in this process is equal to
809349_dedb445343a648779978ba9631419d53.png

A
P0V0R
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B
3P0V0R
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C
9P0V08R
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D
None of these
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Solution

The correct option is C 9P0V08R
This question can be solved by maxima
and minima.
Since the graph shown here is a straight
line, we can write it's equation in
the form of :y=mx+c
here y=P,x=V,m=P2P1V2V1=P02P02V0V0
m=P02V0
c is the intercept and c=3P02
We can find c by extra polating the graph
Now P=P02V0V+3P02 ...(1)
Also, from equation of state, we know the
PV=nRT
for 1mole
PV=RT
or P=RTV
Substituting this in equation (1)
RTV=P02V0V+3P02
T=P02RV0.V2+3P02RV
To find the minimum temperature, we
will differentiate the above equation
i.e., dTdV=2P02RV0.V+3P02R
for dTdV=0P0RV0.V=3P02R
V=3V02
This is the critical point.
d2TdV2=P0RV0
which is negative since
pressure and volume can't
be negative.
pt. of maximum
Now at V=3V02
maximum temperature =Tmax=P02RV09V204+3P02R.3V02
Tmax=9P0V08R
So, option (c) is correct option.

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