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Question: What is the relation between \[{C_P}\] and \[{C_V}\] for methanol? A.\({C_P} > > {C_V}\) B.\({C...

What is the relation between CP{C_P} and CV{C_V} for methanol?
A.CP>>CV{C_P} > > {C_V}
B.CPCV{C_P} \geqslant {C_V}
C.CVCP{C_V} \geqslant {C_P}
D.CV>>CP{C_V} > > {C_P}

Explanation

Solution

CP{C_P} in the system is the amount of heat energy released or absorbed by a unit mass of the substance with the change in temperature at a constant pressure. CV{C_V} is the amount of heat released/absorbed per unit mass of a substance.

Complete answer:
According to the first law of thermodynamics:
ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W, ΔQ\Delta Q is the amount of heat that is given to the system, ΔU\Delta U is the change in internal energy and ΔW\Delta W is the work done.
We can write: ΔQ=ΔU+PΔV\Delta Q = \Delta U + P\Delta V as, ΔW=PΔV\Delta W = P\Delta V
Since ΔQ=nCPΔT\Delta Q = n{C_P}\Delta T and ΔU=nCVΔT\Delta U = n{C_V}\Delta T
Therefore, nCPΔT=nCVΔT+PΔVn{C_P}\Delta T = n{C_V}\Delta T + P\Delta V
We can put the PΔVP\Delta V value from the ideal gas as nRΔTnR\Delta T, the equation will be:
nCPΔT=nCVΔT+nRΔTn{C_P}\Delta T = n{C_V}\Delta T + nR\Delta T
nCPΔT=nΔT(CV+R)n{C_P}\Delta T = n\Delta T({C_V} + R)
CPCV=R{C_P} - {C_V} = R (Where RR is universal gas constant)
At ambient pressure and temperature the isobaric specific heat, CP{C_P} of liquid methanol is 2.53[KJ/KgK]2.53[KJ/KgK], while the isochoric specific heat, CV{C_V} is 2.12[KJ/KgK]2.12[KJ/KgK]. The relation between heat capacities is: CPCV{C_P} \geqslant {C_V}, as the value is approximately the same so they can be equal.
Hence option B is the correct answer.

Note:
The isobaric is used for the substance, when there is a constant pressure, CP{C_P} is an isobaric specific heat because the heat is calculated under the constant pressure, whereas the isochoric is used for the substance, where there is constant volume. CV{C_V} is an isochoric specific heat because the heat is calculated under the constant volume.