Ideal Gases: Isobaric Process

Ideal Gases Isobaric Processes

When examining our favourite ideal gas law of $PV=nRT$, we are going to consider a case where the pressure remains constant. Such a case is referred to as an isobaric process, and $P_1 = P_2$.

For example, you have a balloon, with some initial pressure, volume, and temperature. Say that you leave the balloon outside in the cold Montréal winter overnight. In the morning, you would probably see that the balloon shrank. What happened? Well, the atmospheric pressure remained constant, but the temperature changed, dropping to some infernal degree in the negatives. So since the pressure and moles can be eliminated from the ideal gas equation, we have the relationship $V\propto T$. As the temperature decreased overnight, so must the volume to compensate.

Now, enough of that. Let's investigate further to see what happens to the PV work in an isobaric process:

$$w_{rev} = -\int_{V_1}^{V_2}P_1dV
$$ $$w_{rev}= -P_1(V_2-V_1)=+P_1(V_1-V_2)
$$ $$w_{rev}=P_1(RT_1/P_1-RT_2/P_1)
$$ $$w_{rev}=R(T_1-T_2)
$$

 Note that $V_1>V_2$, and also $n=1$.

 The heat is given by $heat=q_p =\int_{T_1}^{T_2}C_Pd = C_p(T_2-T_1)$

 The change in internal energy is given by 
$$\Delta{U}=q_p+w$$ $$\Delta{U}=C_p(T_2-T_1)-R(T_2-T_1)$$ $$\Delta{U}=(C_p-R)(T_2-T_1)$$ $$\Delta{U}=C_v(T_2-T_1)$$

 In summary for an isobaric process, $$q_p=C_p(T_2-T_1)$$ $$w_{rev}=-R(T_2-T_1)$$ $$\Delta{H} = q_p = C_p(T_2-T_1)$$ $$\Delta{U}=C_v(T_2-T_1)$$

 This process occurs on two different isotherms.