Work done under a constant pressure.

I know $$\Delta U = \Delta V \times \pi_T + C_V\Delta T $$

I thought I can give a try to find work done from above equation under constant pressure for an ideal gas.

$$q_p + w_p = C_V\Delta T$$ $$\implies C_p\Delta T + w_p = C_V \Delta T$$

$$\implies w_p = -\Delta T (C_p - C_V)$$ $$\implies w_p = - R\Delta T $$

Adiabatic work is $w_\mathrm{ad} = C_V \Delta T$

For a adiabatic work under constant pressure $w_\mathrm{ad} = w_{q} \implies C_V = -R \implies C_p = 2R$

Constant $C_p, C_V$ and negative $\gamma$ does not look correct to me.

  • Which of my steps are incorrect ?

1 Answer 1


You can't do adiabatic reversible work at constant pressure. The pressure has to change. If you want to do adiabatic irreversible work at constant pressure, say by suddenly increasing or decreasing the pressure load to begin with, and then holding it constant during the process, this is sometimes referred to as adiabatic work at constant pressure, but it really isn't. The final pressure will not be equal to the initial pressure that was present before you changed the applied load.

  • $\begingroup$ So I should use Kirchoff's law to determine $q_p$ ? $\endgroup$
    – user31607
    Feb 13, 2017 at 11:03
  • $\begingroup$ I also did not understand why work can't be done on a adiabatic system under constant pressure. $\endgroup$
    – user31607
    Feb 13, 2017 at 11:45
  • $\begingroup$ In a closed system, the only way to compress or expand a gas such that the initial pressure is equal to the final pressure, and is constant along the entire path is to raise its temperature by adding heat. But that's not adiabatic. Try solving a reversible adiabatic path, and see how that plays out. $\endgroup$ Feb 13, 2017 at 12:42
  • $\begingroup$ Ok I will try that out. Is my statement, $w_p = - R\Delta T$, correct for work done under constant pressure for an ideal gas ? $\endgroup$
    – user31607
    Feb 13, 2017 at 13:23
  • $\begingroup$ Yes, if that's the work done per mole by the surroundings on the gas. Don't forget that, in an adiabatic reversible process, the pressure is changing, so you have to take the integral of PdV=RTdV/V to get the work. $\endgroup$ Feb 13, 2017 at 13:37

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