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Question: The process \(CD\) is shown in the diagram. As the system is taken from \(C\) to \(D\), what happens...

The process CDCD is shown in the diagram. As the system is taken from CC to DD, what happens to the temperature of the system?

A. Temperature first decreases and then increases
B. Temperature first increases and then decreases
C. Temperature decreases continuously
D. Temperature increases continuously​

Explanation

Solution

With the data given to us, we can easily calculate the conditions at both the ends of CDCD using the ideal gas relations. However, to find what happens during the process, we can arbitrarily choose a point, such as the midpoint of CDCD and evaluate the conditions at that point.
Formulas used: PV=nRTPV = nRT
Where PP is the pressure, VV is the volume, nn is the number of moles, RR is the universal gas constant and TT is the absolute temperature.

Complete step by step answer:
-From the ideal gas law, we have:
PV=nRTT=PVnRPV = nRT \Rightarrow T = \dfrac{{PV}}{{nR}}
Where PP is the pressure, VV is the volume, nn is the number of moles, RR is the universal gas constant and TT is the absolute temperature.
-At point CC, we have P=3p0P = 3{p_0} and V=v0V = {v_0}
Substituting these values, we have:
TC=3p0v0nR{T_C} = \dfrac{{3{p_0}{v_0}}}{{nR}} ………………… (1)
Where TC{T_C} is the temperature at the point CC
-At point DD, we have P=p0P = {p_0} and V=3v0V = 3{v_0}
Substituting these values, we get:
TD=3p0v0nR{T_D} = \dfrac{{3{p_0}{v_0}}}{{nR}} ……………..…. (2)
Where TD{T_D} is the temperature at the point DD
Now let us evaluate the temperature at the midpoint of CDCD.
-As we are choosing the midpoint, both pressure and volume at the midpoint will be the average of pressure and volume at the two ends of CDCD
-Therefore, at the midpoint, P=3p0+p02P = \dfrac{{3{p_0} + {p_0}}}{2}
On solving this, we get:
P=4p02=2p0P = \dfrac{{4{p_0}}}{2} = 2{p_0}
Similarly, volume at the midpoint, V=v0+3v02V = \dfrac{{{v_0} + 3{v_0}}}{2}
On solving this, we get:
V=4v02=2v0V = \dfrac{{4{v_0}}}{2} = 2{v_0}
Substituting these values, we get:
Tmidpoint=2p0×2v0nR=4p0v0nR{T_{{{midpoint}}}} = \dfrac{{2{p_0} \times 2{v_0}}}{{nR}} = \dfrac{{4{p_0}{v_0}}}{{nR}} ………………… (3)
From equations (1), (2) and (3) we can clearly see that temperatures at CC and DD are equal and the temperature at the midpoint is higher than either of these.
Hence, temperature first increases from CC till the midpoint and then decreases till DD.

Therefore, the correct option is B.

Note: Note that we have taken the conditions at the midpoint as the average value of both the ends of CDCD only because the process CDCD is seen to be a linear function, that is, a straight line. If the shape of the graph were to be different, we would have to use integral methods to evaluate the conditions at the midpoint. The initial increase in temperature is due to the fact that the pressure is continuously reducing and the volume is continuously increasing, till they reach a maximum at the midpoint. After the midpoint, the fall in pressure is drastic and this leads to decrease in temperature.