# Work done in adiabatic process

When we try to establish a relation between the pressure and temperature in adiabatic process we come across a equation..

$dU = dq - PdV$

$dq=0$ (Adiabatic process) and,

$dU=C_v.dT$ (Heat capacity at constant volume)

Therefore, $C_v.dT = -PdV$$\tag1 In this equation we are using heat capacity defined at constant volume but their still is some work done by the system (i.e, PdV is not 0 or dV not equals to 0). The first part of the equation (1) is implying that the volume is constant but the second part is implying that the volume is not constant (if it was there would be no work done). Then why there is this contradiction ? ## 4 Answers Even though we call C_v the heat capacity at constant volume, what we really mean by the subscript v is that this is the way we measure C_v. At constant volume, we can determine the heat capacity of the material by measuring the heat transferred dQ=dU=C_vdT. But this same heat capacity also applies to all other situations for an ideal gas if we recognize that, for an ideal gas, U depends only on temperature, such that dU=C_vdT. It is just that, in these other situations (involving work), dQ is not equal to C_vdT. • According to this , since you say "heat capacity is same for all situations" (quoted from your answer) then Cp should be equal to Cv since Cp becomes one of the situation where volume is changing. But this disagrees with Cp-Cv=R. I'm a little too confused at this point. Could you please explain in better terms? – user29660 Commented Apr 16, 2017 at 10:19 • C_p and C_v are not the same, and, in thermodynamics, neither of them is defined in terms of dQ. The proper definitions in thermodynamics are C_p=(\partial H/\partial T)_p and C_v=(\partial U/\partial T)_v. For an ideal gas (whose internal energy and enthalpy depend only on temperature), dU=C_vdT and dH=C_pdT for any arbitrary process path. Commented Apr 16, 2017 at 11:12 • Would you mind editing the answer here ? I like the explanation in your new post. (+2) Commented Sep 19, 2017 at 20:51 • At different times, I've tried to explain this in different ways. Some work better than others. For some reason, this is a very difficult concept to explain. There are other posts I've written that provide even more detail. Since this is from almost a year ago, I don't think many people go back to it. I'm unmotivated to go back now and make changes. Commented Sep 19, 2017 at 22:13 I will just expand on what Chester Miller had already said. The molar specific heat at constant volume C_v is basically the amount of heat required to raise the temperature of 1 mol of the gas by 1 K while keeping the volume constant. When we apply heat to an ideal gas, some of that heat is used to increase the internal energy U of the gas and the rest is eventually used up to do the work to increase its volume. (Note that increasing U essentially means increasing the temperature since U is a function of T for any ideal gas.) Now if we were to keep the volume constant somehow, then the total applied heat would be used only to increase the internal energy or so to speak the temperature and nothing else. So a different way to define C_v is to say, it is the amount of heat required to raise the temperature of 1 mol of the gas by 1 K, where the said heat is used to only to change the internal energy U or so to speak the temperature and nothing else. Now let's consider a case where change in volume is allowed. Say we applied dQ amount of heat to the a gas. Now we know some of that heat is going to account for the change in internal energy, let's say that amount is dQ_u. And the rest of the heat will eventually account for the change in volume, and say that is dQ_w. So instead of writing dQ = dU + dW we write,$$dQ = dQ_u +dQ_w$$So we can see that the heat dQ_u is only used to change the internal energy and not the volume and so dQ_u = dU and we can say according to the definition of C_v the following$$C_v = \frac{dQ_u}{mdT}\ \Rightarrow\ dQ_u = mC_vdT\ \Rightarrow\ dU=mC_vdT$$So, you see dU = mC_vdT whether or not there is any change in volume. The same argument goes for the adiabatic process you mentioned. • Greatly done. Thanks for giving some time to this question. Commented Jan 8, 2018 at 7:43 The answer is pretty simple. We will prove it for an ideal monoatomic gas, the proof can be adjusted for other gasses as well. Firstly, I will use the convention that $$\Delta E_{int} = Q + W$$ where the sign of $$Q$$ is positive if heat is transferred to the system and the sign of $$W$$ is positive if the positive work is done to the system (by the environment). Note that we have $$E_{int} = \frac{3}{2}nRT$$ for a monoatomic gas, and thus $$\Delta E_{int} = \frac{3}{2} nR\Delta T.$$ What this essentially means is that regardless of the process, the change of the internal energy depends only on the final and initial temperature. If the volume of the gas is constant, then $$W = 0$$, so $$Q = \Delta E_{int}$$. By the definition of $$C_v$$, we have $$C_v = \frac{Q}{n\Delta T} =\frac{\Delta E_{int}}{n\Delta T}$$ $$\implies \Delta E_{int} = C_v n \Delta T.$$ Now equating the two equations give $$C_v = \frac{3}{2}R$$. Note that $$\Delta E_{int} = \frac{3}{2} nR\Delta T$$ applies in any process, and thus $$\Delta E_{int} = nC_v dT$$ also applies in any process since they are the SAME equation, and that $$C_v$$ is independent of the type of the monoatomic gas. Thermal capacity at constant volume \textrm C_V is defined as$$\mathrm C_V ~=~\left(\frac{\partial U}{\partial T}\right)_V$$where U is the internal energy of the system. For an ideal gas, U=U(T); so,$$\mathrm C_V~\mathrm dT ~=~ \mathrm dU\,.\tag I$$Substituting \mathrm{(I)} in the First Law of Thermodynamics,$$\mathrm C_V~\mathrm dT +đw~=~ đQ\tag{II} $$From which, for an adiabatic process,$$\mathrm C_V~\mathrm dT ~=-~P~\mathrm dV,$$which is actually$$\mathrm dU ~= -~P~\mathrm dV\,.$\$

Nothing is contradictory here.

• In your second line(the first equation) you've taken (dU/dT) at constant volume (you have specified using subscript 'v' which is why you get Cv). Now since volume is defined to be a constant, dV will be 0. Thus giving you work done=0. Isn't that contractory?
– user29660
Commented Apr 16, 2017 at 10:24