An ideal gas having 2 moles (fixed) is subjected to the changes as shown is (P–V) diagram. Select the
correct option(s) from the following diagram
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Answer:
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>>An ideal gas is taken from state A (Pres
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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:
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This question has multiple correct options
Hard
Solution
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Correct options are A) , B) and D)
Work done = Area under P−V graph
As A
1
>A
2
∴W
given process
>W
isothermal process
Thus (A) is correct.
In the given process P−V 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=−aV
2
+bV
⇒nRT=−aV
2
+bV[∵PV=nRT(From ideal gas equation)]
⇒T=
nR
1
(−aV
2
+bV)
The above equation is of parabola.
Thus (B) is correct.
T=
nR
1
(−aV
2
+bV)
dV
dT
=0
⇒
nR
1
(−2aV+b)=0
⇒−2aV+b=0
⇒V=
2a
b
Now,
dV
2
d
2
T
=−2a=−ve
∴T has some maximum value.
Now, from ideal gas equation,
T∝PV
∵(PV)
A
=(PV)
B
∴T
A
=T
B
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.
Answer:
Correct options are A),B) and D)
Explanation:
Work done=Area under P--V graph
As A1>A2
Therefore, W given process>W isothermal process
Thus A is correct
In the given process P--V equation will be a straight line with negative and positive intercept,
P= -aV+b
nRT= -aV²+bV
T=1/nR (-aV²+bV)
Eqn is parabola.
Thus B is correct.
T=1/nR(-aV²+bV)
dT/dV=0
1/nR(-aV²+bV)=0
V=b/2a
d²T/dV²=-2aV+b=-ve
T has max value.
From ideal gas equation, T∞PV
(PV)A=(PV)B
TA=TB
We conclude that temperature are same at A and B and also have maxi value in between them.
Hence in going from A and B, the temperature will first increase to maxi value and then decrease. Thus (D) is correct.
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