91Ó°ÊÓ

$(T_2+273)K=\frac{200kPa}{500kPa}×1×(400+273)K$

We have

$(T_2+273)K=\frac{200kPa}{500kPa}×1×(400+273)K$

From the ideal gas theory pV = nRT

An isothermal process, during which the system's temperature remains constant. An isobaric process, during which the system's pressure does not change. An isochoric process, during which the system's volume does not change.

The problem could be

According to the ideal gas law, pressure is proportional to temperature. As a result, while the volume is constant, the above equation indicates the relationship between pressure and temperature. "A gas in a hard container is at a temperature of 400^oC and a pressure of 500 kPa . The container's internal pressure is decreased to 200 kPa. Calculate the final temperature in degrees Celsius."

The diagram for the cycle can be drawn as below

Solve the given equation

$$\begin{gathered} (Tâ‚‚+273)K=\frac{200kPa}{500kPa}\left(1\right)(400+273)K \\ =-4^{o}C \end{gathered}$$