Q. CP is always greater than CV for a gas. Which of the following statements provide, partly or wholly, the reason for this??
a) No work is done by a gas at constant volume
b) When a gas absorbs heat at constant pressure, its volume must change.
c) For the same change in temperature, the internal energy of a gas changes by a smaller amount at constant volume than at constant pressure.
d) The internal energy of an ideal gas is a function only of its temperature.
Plz. tell me the answer to this question with good reasoning.
Dear Ankul,
The reasoning for UR ques.is as follows:-
When a gas is heated, both its vol. & pressure change. Let me estimate the amt. of heat required by gas to heat it thro' 1oC, when either its vol. or pressure is kept const.
When vol.is kept const.-
Pressure on the gas is increased so that its vol. tremains const. As the gas can't perform work (V const.) the heat supplied will increase only the temp. or int.energy of the gas.
Therefore, in case of Cv heat is required only for incresing temp.of gas thro' 1oC.
In view of the above pt. option a is correct.
When pressure is kept const.-
When gas is heated at const. pressure, it expands also.Therefore, the heat supplied at const. pressure will partly increase its int. energy & partly will be utilised in performing work against ext.pressure.Therefore, in case of Cp more heat as compared to Cv will be required for increasing temp.of gas thro' 1oC.
Hence,Cp > Cv
In view of the above pts.options a,b,c seem to be correct to me.
The reasoning for UR ques.is as follows:-
When a gas is heated, both its vol. & pressure change. Let me estimate the amt. of heat required by gas to heat it thro' 1oC, when either its vol. or pressure is kept const.
When vol.is kept const.-
Pressure on the gas is increased so that its vol. tremains const. As the gas can't perform work (V const.) the heat supplied will increase only the temp. or int.energy of the gas.
Therefore, in case of Cv heat is required only for incresing temp.of gas thro' 1oC.
In view of the above pt. option a is correct.
When pressure is kept const.-
When gas is heated at const. pressure, it expands also.Therefore, the heat supplied at const. pressure will partly increase its int. energy & partly will be utilised in performing work against ext.pressure.Therefore, in case of Cp more heat as compared to Cv will be required for increasing temp.of gas thro' 1oC.
Hence,Cp > Cv
In view of the above pts.options a,b,c seem to be correct to me.
Well Sooraj,
c is definitely wrong as it is a wrong statement. Under any condition, Internal energy changes by same amount i.e. nCvT. So, answers should be a,b i suppose. Thnx for ur good reply.
c is definitely wrong as it is a wrong statement. Under any condition, Internal energy changes by same amount i.e. nCvT. So, answers should be a,b i suppose. Thnx for ur good reply.
Considering how vaguely the options a) & b) seem to indicate that CP is greater than CV i fail to see how option d) is not right too.
Option d is right and i thought of a simple proof.Since typing it is difficult, i searched and found another proof . It is on http://pruffle.mit.edu/3.00/Lecture_11_web/node1.html
This one simply proves that Cp- Cv = R and hence says Cp>Cv
This one simply proves that Cp- Cv = R and hence says Cp>Cv
Well, even i couldn't see nything wrong in this proof. But, in almost all the books the answer to this ques. is (a) and (b) only.
Well, Ankul,
This problem might have been a conventional one and might have been asked long before.
So, Many books might have taken this question and hence they have same answers.But, I am quite sure that d is also a correct option.
This problem might have been a conventional one and might have been asked long before.
So, Many books might have taken this question and hence they have same answers.But, I am quite sure that d is also a correct option.
well yes, even i can't think of why d is wrong.
a,b,d are correct. work is not done as deltaV is zero for a
if deltaP not equal to 0 then deltaV cannot be zero so b is also correct and since deltaU=nCvdeltaT d is also correct.
if deltaP not equal to 0 then deltaV cannot be zero so b is also correct and since deltaU=nCvdeltaT d is also correct.
Well Ankul,
a,b are certainly correct.U being state function I can change the temp through any process and the resulting change in U will be same.So simplest change i can think of is const. volume in which case (deltaU)=nCv(deltaT)