Q. 8.38 - An Introduction to General, Organic, and Biological Chemistry

Question

A gas sample has a pressure of $2.35 atm$ when the temperature is  $-15°C$. What is the final pressure, in atmospheres, when the temperature is $46°C$, with no change in the volume or amount of gas?

Step-by-Step Solution

Answer

The final pressure of the gas is $2.9 atm$.

Step 1: Given Information

We need to find the final temperature of the gas. 

Step 2: Simplify

Considering the Gay- Lussac's Law that if a constant volume and amount of gas are maintained, the pressure will be increased. In the temperature-pressure relationship which is known as Gay-Lussac's law, the pressure of a gas decreases, and a decrease in temperature decreases the pressure of the gas till the volume of the gas doesn't change. i.e.   

$\frac{P_1}{T_1}=\frac{P_2}{T_2} ... \left(1\right)$

Step 3: Calculation

Here we have the quantities available, that is   

the initial temperature $T_1=-15°C=\left(-15+273\right)K=258K$

initial pressure $P_1=2.35 atm$

the final temperature  $T_2=46°C=\left(46+273\right)K=319K$

Now, calculating the final pressure $P_2$ from the equation  $\left(1\right)$

$$\begin{gathered} \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ {P}_2=\frac{P_1\times T_2}{T_1} \end{gathered}$$

Putting the values: 

$$\begin{gathered} P_2=\frac{2.35 atm\times 319K}{258K} \\ {P}_2=\frac{749.65 atm}{258} \\ =2.9 atm \end{gathered}$$

The final pressure is  $2.9 atm$.