Question

$0.020mol$ of gas undergoes the process shown in FIGURE EX18.37.

a. What type of process is this?

b. What is the final temperature in $°C$?

c. What is the final volume $V_2$?

Short answer

a. This type of process is called Isothermal.

b. The final temperature is $-29°C$?

c. The final volume is$133c{m}^3$?

Step-by-step solution

: Given information and Formula used 

Given : Gas's initial temperature : $900°C$

Undergoes the process shown in FIGURE EX18.36

Theory used :

In Thermodynamics, an Isothermal Process is a type of thermodynamic process in which the temperature $T$of a system remains constant, i.e., $\Delta T=0$.

The ideal gas law ($PV=nRT$) relates the macroscopic properties of ideal gases. It shows the relationship between the four state variables of an ideal gas.

 Defining what type of process this is .

(a) For the isothermal process, the $PV$ diagram is shown as a hyperbolic graph, as can be seen in FIGURE EX18.37.

Hence, the process isIsothermal .

Calculating the final temperature in °C

(b) We solve equation $PV=nRT$for $T$ to be in the form $T=\frac{PV}{nR}$

Given the initial pressure is $p_1=1atm=1.013x{10}^{5}Pa$and the initial volume $V_1=400c{m}^3$.

So, the final temperature will be :

role="math" localid="1648277408321" $$\begin{gathered} T=\frac{P_1{V}_1}{nR} \\ =\frac{(1.013x{10}^{5}Pa)(400x{10}^{-6}{m}^3)}{(0.020mol)(8.314J/mol.K)} \\ =244K \end{gathered}$$

Androle="math" localid="1648277445429" $244K=\mathbf{-}\mathbf{29}\mathbf{°}\mathit{C}$

Finding the final volume . 

(c) The constant temperature process for the closed system has the constant product$pV$. That is
$$\begin{gathered} p_1{V}_1=p_2{V}_2 \\ \Rightarrow V_2=\left(\frac{p_1}{p_2}\right)V_1 \\ =\left(\frac{1atm}{3atm}\right)400c{m}^3 \\ =\mathbf{133}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{3}} \end{gathered}$$