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As pressure is increased, do we require more energy to increase the temperature from given temperature for the same mass. For example, if we heat water at 1 atm to raise its temperature by 20 degree Celsius, will the heat required be the same when pressure is 3 atm?

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I shall try to explain using first law of thermodynamics and Maxwell' relations. For simplicity I am considering $C_v$ . You may extend the argument to $C_p$ as well. Let us consider the change in $C_v$ with w.r.t $V$. From the definition of $C_v$ we can write,

$\left(\frac{dC_v}{dV}\right)_T = \left[\frac{d}{dV}\left(\frac{dU}{dT}\right)_V\right]_T ...(1)$

Switch order of differentiation and from the first law of thermodynamics we can write

$\left[\frac{d}{dT}\left(\frac{dU}{dV}\right)_T\right]_V = \left[\frac{d}{dT}\left(T \frac{dS}{dV} - P \right)_T\right]_V .. (2)$

From Maxwell's relation we have

$ \left(\frac{dS}{dV}\right)_T = \left(\frac{dP}{dT}\right)_V .. (3) $

Substitute Eqn $(3)$ in $(2)$ and after applying product rule, you may obtain

$\left(\frac{dC_v}{dV}\right)_T = T \left(\frac{d^2P}{dT^2}\right)_V$

For an ideal gas or a gas having an equation of state of the form $PV = kT$ or a linear function of $T$, the second derivative of $P$ w.r.t $T$ vanishes and therefore no dependence. For any other equation of state, this is the dependence of specific heat w.r.t to pressure or volume. You may work out in a similar fashion for $C_p$ w.r.t $P$. Only difference is that the thermodynamic quantity will be Enthalpy and not Internal Energy

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For water and most solids/liquids, yes but very slightly. When you heat the water it expands, which does work against the surrounding pressure. At higher pressure, the expansion takes more work. For example, in heating water from $10 C$ to $30 C$, the density decreases from $999.7026 \frac {kg}{m^3}$ to $995.6502 \frac {kg}{m^3}$, so $1 kg$ of water expands by $2.066\cdot10^{-6} m^3$. At $1 atm= 101325 \frac N{m^2}$ this does $0.209 J$ of work and at $3 atm$ it would be three times this or $0.627 J$. By comparison, the energy to heat the water due to the temperature is about $20000*4.183=83680 J$ If you care about this level of accuracy, you need to see what conditions were used in your tables, and there may be other effects to consider. The difference over pressure should be reliable.

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  • $\begingroup$ Just to be clear, I believe this argument would apply to Cp, but would be ineffective regarding Cv. That doesn't mean that it doesn't change, and I believe it is proven in another answer. But this answer is a direct and clear answer for Cp. $\endgroup$ – Alan Rominger Jul 9 '14 at 22:21
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Look at Cp and Cv. One is specific heat or heat capacity at constant pressure the other one at constant volume. These are constant for ideal or perfect gases and the approximations can be looked up in tables. For liquids it gets rather messy, so you should assume it stays constant, or if you need to be accurate you could look this up in tables such as the IF 67 or consult a program such as fluidprop to get accurate values for matlab calculations. Unless you are messing about near the boiling point, with large pressure changes or have some strange fluids, assume it is constant, the errors due to heat loss and other factors will be far larger.

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