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Question

An ideal gas is taken from state A (Pressure P, Volume V ) to the state B (Pressure P/2, Volume 2V) along a straight line path in PV diagram as shown in the adjacent figure.
Select the correct statement (s) among the following:
1059596_edf80973980d4aba9cdf4eed78a27938.png

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
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B
in the T-V diagram, the path AB become part of parabola
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C
in the P-T diagram, the path AB become part of hyperbola
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D
in going from A to B, the temperature T gas of the gas first increases to a maximum values then decreases
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Solution

The correct options are
B 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
C in the T-V diagram, the path AB become part of parabola
D in going from A to B, the temperature T gas of the gas first increases to a maximum values then decreases
Work done = Area under PV graph
As A1>A2
Wgiven process>Wisothermal process
Thus (A) is correct.
In the given process PV equation will be a straight line with negative slope and positive intercept, i.e.,
P=aV+b
Here, a and b are positive constants.
P=aV+b
PV=aV2+bV
nRT=aV2+bV[PV=nRT(From ideal gas equation)]
T=1nR(aV2+bV)
The above equation is of parabola.
Thus (B) is correct.
T=1nR(aV2+bV)
dTdV=0
1nR(2aV+b)=0
2aV+b=0
V=b2a
Now,
d2TdV2=2a=ve
T has some maximum value.
Now, from ideal gas equation,
TPV
(PV)A=(PV)B
TA=TB
We conclude that temperatures are same at A and B but in between the temperature has a maximum value.
Hence in going from A to B, the temperture will first increase to a maximum value and then decrease.
Thus (D) is correct.

1088823_1059596_ans_19a922c9cb674300841c51e4052eafc2.png

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