91Ó°ÊÓ

The process that takes place at constant pressure is called an isobaric process. Using the formulas for work done W in an isobaric process, and the first law of thermodynamics, we can find the change in the internal energy${\mathbf{Δ·¡}}_{\mathrm{int}}$of the gas. Also, by using theideal gas law in the ratio form, we can find the final temperature ${\mathbf{T}}_2$of the gas.

${\mathbf{Δ·¡}}_{\mathrm{int}}\mathbf{=}\mathbf{Q}\mathbf{-}\mathbf{W}$

1. The pressure.$p=25{\text{ N/³¾}}^{\text{2}}$
2. The final volume$V_f=1.8{\text{m}}^3$
3. The initial volume.$V_i=3.0{\text{m}}^3$
4. The initial temperature$T_1=300\text{K}$
5. In the process, the energy lost by the gas as heat is$Q=−75\text{J}$ .

$W=p\mathrm{Δ}V$

Where p is the pressure and $\mathrm{Δ}V=(V_f−V_i)$is change in the volume

Therefore,

$\begin{array}{c}W=p(V_f−V_i)\\ =25\text{ N}/{\text{m}}^2×(1.8{\text{″¾}}^3−3.0{\text{″¾}}^3)\\ =−30\text{ J}\end{array}$

From the first law of thermodynamics, we have-

$\mathrm{Δ}{E}_{\mathrm{int}}=Q−W$

Where Q is the heat transferred and$\mathrm{Δ}{\text{E}}_{int}$is the change in the internal energy o the gas. It is given that Q amount of energy leaves the system; therefore, we can consider Q as negative.$Q=−75\text{ J}$

$\begin{array}{c}\mathrm{Δ}{E}_{\mathrm{int}}=−75\text{ J}−(−30\text{ J})\\ =−75\text{ J}+\left(30\text{ J}\right)\\ =−45\text{ J}\end{array}$

Internal energy is decreased by$−45\text{ J}$

The change in internal energy of the gas is $-45J$and the final temperature of the gas is $180\text{‿é}$.