Isobaric Process- Constant Pressure

  • Calculate total work done in a constant pressure thermodynamic process.

Isobaric Process

A process in which pressure remains constant.
A constant pressure process is a horizontal path in the P-V diagram - right for expansion and left for compression. Example: a gas in a container sealed with a freely-sliding massive piston.

The work done during gas expansion is: \[dW = - P dV\]

\[W = \int_1^2 p dV = p \int_1^2 dV = p(V_2 - V_1) = mR (T_2 - T_1)\]

\[Q = \int_1^2 mC_P dT = m C_P (T_2 - T_1)\]

By Applying first law of thermodynamics, \[Q - W = \Delta U\] \[m C_P (T_2 - T_1) - mR (T_2 - T_1) = m C_V (T_2 - T_1)\]

\[C_P - C_V = R\]

This is also known as Meyer’s formula.

Solved Examples

Solved Example:


Addition of heat at constant pressure to a gas results in:

Correct Answer: D

Solved Example:


A piston/cylinder device contains one kilogram of a substance at 0.8 MPa with a specific volume of 0.2608 m$^3$ /kg. The substance undergoes an isobaric process until its specific volume becomes 0.001115 m$^3$/kg. Find the total work done in the process.

\[P_{1} =0.8\times 10^{6} Pa, V_{1} =0.2608 \dfrac {m^{3}}{kg}, V_{2} =0.001115 \dfrac {m}{kg}\] \[W = PdV =0.8\times 10^{6}\times \left[ 0.001115-0.2608\right] =207.7\ kJ\]

Correct Answer: B

Solved Example:


When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas, is:

\[ f = \dfrac{\Delta U}{\Delta Q} = \dfrac{m C_v \Delta T}{m C_p \Delta T} = \dfrac{C_v}{C_p} = \dfrac{1}{\gamma}\] For ideal diatomic gas, $\gamma$ = $\dfrac{7}{5}$ \[f = \dfrac{5}{7}\]

Correct Answer: D