Question

The temperature increases from 300 K to 1200 K as a gas undergoes

the process shown in FIGURE Q18.11. What is the final

pressure?

Short answer

The final pressure of gas = 8 atm

Step-by-step solution

Visualisation

The ideal gas equation can be used in this problem.

In the ideal gas equation,

$PV=nRT$

Here, localid="1648461240466" $P,V,n,R\&T$are respectably pressure, volume, number of moles, universal gas constant and absolute temperature of gas.

In the given problem, $V,n\&R$are constants.

So,

localid="1648462419830" $\begin{array}{rcl}P& \propto & T\\ \frac{P_2}{P_1}& =& \frac{T_2}{T_1}\\ P_2& =& \left(\frac{T_2}{T_1}\right)P_1\end{array}$

Here, $P_1\&T_1$are the initial pressure & temperature and $P_2\&T_2$are the final pressure & temperature.

Step: 2  Calculation of the final pressure of gas

As given in the question,

$$\begin{gathered} T_1=300K, \\ {T}_2=1200K, \\ {P}_1=2atm \end{gathered}$$

Then,

$P_2=\left(\frac{T_2}{T_1}\right)P_1=\left(\frac{1200K}{300K}\right)\left(2atm\right)=8atm$

So, 8 atm is the final pressure of gas.