### Isobaric Process- Constant Pressure

• Calculate total work done in a constant pressure thermodynamic 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 Example:

#### 72-3-01

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

### Solved Example:

#### 72-3-02

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.

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

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