# Why is specific heat capacity at constant pressure not equal to the rate of change of internal energy

The formula for internal energy is $U= nC(v)dT$. But why cant we use U=nC(p)dt where C(p) is the specific heat at constant pressure

Someone might say that we can use $C(v)$ because Internal energy of an ideal gas does not depend on the volume so even if volume is changing we can use $C(v)$. But then the internal energy of an ideal gas is also not dependent on the pressure. Its only depended on the temperature of the gas. So why cant we use $C(p)$?

Also for real gasses both $C(v)$, $C(p)$ shall be correct. Isn't it?

• – Paul T. Mar 2 '17 at 16:23

In Thermodynamics, the specific heat capacity at constant volume and the specific heat capacity at constant pressure are physical properties of a material (irrespective of process) that are precisely defined as follows: $$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$ and $$C_p=\left(\frac{\partial H}{\partial T}\right)_P$$where the specific internal energy U (per mole or per unit mass) and the specific enthalpy H (per mole or per unit mass) are also physical properties of the material (irrespective of process), with $$H=U+PV$$where V is the specific volume (per mole or per unit mass). For an ideal gas, U and H are functions only of temperature, and not pressure or volume. Therefore, for an ideal gas $$C_p=\left(\frac{dH}{dT}\right)=\frac{dU}{dT}+\frac{d(PV)}{dT}=C_v+\frac{d(RT)}{dT}=C_v+R$$

Problem 1: This is a 2 step process involving 1 mole of ideal gas. In Step 1, the system starts out at $P_1$, $V_1$, and $T_1$, and is heated at constant volume to temperature $T_2$ (and corresponding pressure $P_2$). In Step 2, the gas is allowed to expand isothermally and reversibly (being held in contact with a constant temperature bath at $T_2)$ until the pressure is again $P_1$.

Step 1 Questions:

In terms of $T_1$, $T_2$, and $P_1$, what is the final pressure $P_2$ at the end of Step 1?

In terms of $T_1$ and $T_2$, what is $\Delta (PV)$ in Step 1?

In terms of $T_1$ and $T_2$, what is $\Delta U$, $\Delta H$, W, and Q in Step 1?

Step 2 Questions:

In terms of $T_1$, $T_2$, and $V_1$, what is the final volume at the end of Step 2?

What is $\Delta (PV)$ in Step 2?

In terms of $T_1$ and $T_2$, what is W and Q in Step 2?

What is $\Delta U$ and $\Delta H$ in Step 2?

Overall Process Questions:

In terms of $T_1$ and $T_2$, what are Q and W for the overall process?

In terms of $T_1$ and $T_2$, what are $\Delta U$ and $\Delta H$ for the overall process?

What is $\Delta U$ divided by $\Delta T = (T_2-T_1)$ for the overall process?

What is $\Delta H$ divided by $\Delta T = (T_2-T_1)$ for the overall process?

• But in an adaibatic process we take U= C(v)dT where Q=0 and W is not equal to zero. Volume is changing and dV is not 0. Also P is changing. Why don't we use C(p)dT here. – E2n Mar 5 '17 at 7:03
• Because the first law of thermodynamics calls for the use of U, not H. Of course, if the process is also reversible, one could also use either $dU=C_vdT=-PdV$ or $dH=C_pdT=VdP$. The two equations are equivalent. You are aware that, for an ideal gas, U and H are functions only of T (and not P or V), so dU is always equal to CvdT and dH is always equal to CpdT (irrespective of what P and V are doing), correct? – Chet Miller Mar 5 '17 at 12:46
• But when we work at constant volume only then U= Q – E2n Mar 5 '17 at 16:49
• But when we work at constant volume only then U= Q but C(V) is defined as the heat capacity at constant V. When the volume is not constant then its not heat capacity at constant V. Like in JohnRennie's answer he tells why when W=0 then Q=U and only then U=C(v)dT . How can we before hand take U=C(V)dT . It could be anything . It can be C(P) or any value of C. You said that U for ideal gas is independent of V And P. Then if dU/dV = dU/dP =0 Hence C(v) and C(P) both are OK. I am getting confused. Even while calculating Joule Coefficient we use C(V). If the gas was real then also C(V) is right ? – E2n Mar 5 '17 at 17:01
• Let's forget about Cp for the moment. Suppose we hold the volume of an ideal gas constant, and we increase the temperature. Then the change in internal energy is $\Delta U=C_v\Delta T$, right. Then we change the volume to any other value, holding the temperature constant. This can be done by holding the gas in contact with a constant temperature reservoir. For this second step, $Q=W=\int{PdV}$, and $\Delta U=0$ So the overall change in U is still $\Delta U=C_v\Delta T$ Now, since U depends only on T, it doesn't depend on what path we took for get from the initial state to the 2nd state. – Chet Miller Mar 5 '17 at 20:03

The internal energy of a closed system is defined by:

$$dU = \delta Q + \delta W$$

i.e. it is the energy added to the system plus the work done on the system. The work is $-PdV$ giving:

$$dU = \delta Q - PdV$$

If we work at constant volume then $dV = 0$ and the internal energy is then simply the heat added i.e. $C_vdT$. If we work at constant pressure then the heat added is $C_pdT$, but we'd have to calculate the work done by the system as it expanded and then subtract that off to get the internal energy.

• Oh! Thanks . I just did not think about that. Just one thing more . Could you please tell me ( sort examples ) that which relations developed in the thermodynamics that an undergrad studies change due to the faulty assumption of an ideal gas. Obviously 1st & 2nd laws won't change ( since they are laws , but I can't prove it) and I know about van der waals equation. But which general results ( like above) would change due to the gas being real shall change because I feel that certain equations shall change ( I can not express it properly)..... ! Thank you ! – E2n Mar 2 '17 at 18:10
• But in an adaibatic process we take U= C(v)dT where Q=0 and W is not equal to zero. Volume is changing and dV is not 0. Also P is changing. Why don't we use C(p)dT here. – E2n Mar 5 '17 at 7:02