Q. 8.36 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the final temperature, in degrees Celsius, for each of the following, if  $V, n$ do not change:

a. A sample of helium gas with a pressure of $250 Torr$ at $0°C$ is heated to give a pressure of $1500 Torr$.

b. A sample of air at   $40°C$ and $740 mmHg$ is cooled to give a pressure of $680 mmHg.$

Step-by-Step Solution

Answer

a. The final temperature of helium gas  is  $1365°C$.

b. The final temperature of air  is  $14.62°C$.

Part (a) step 1: Given Information

We need to find the final temperature of the helium gas. 

Part (a) 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.  

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

Part (a) step 3: Calculation

Here we have the quantities available, that is  

the initial temperature  $T_1=0°=\left(0+273K\right)=273K$

initial pressure $P_1=250 Torr$

the final pressure $P_2=1500 Torr$

Now, calculating the final temperature $T_2$:

$$\begin{gathered} T_2=\left(T_1\right)\times \left(P_2\right)÷\left(P_1\right) \\ {T}_2=\left(273K\right)\times \left(1500 mmHg\right)÷\left(250mmHg\right) \\ {T}_2=1638 K=\left(1638-273\right)°C=1365°C \end{gathered}$$

Part (b) step 1: Given Information

We need to find the final temperature of the air gas. 

Part (b) 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.  

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

Part (b) step 3: Calculation

Here we have the quantities available, that is  

the initial temperature  $T_1=40°C=\left(40+273\right)K=313K$

initial pressure$P_1=740mmHg$

the final pressure $P_2=680mmHg$

Now, calculating the final temperature $T_2$:

$$\begin{gathered} T_2=\left(T_1\right)\times \left(P_2\right)÷\left(P_1\right) \\ {T}_2=\left(313K\right)\times \left(680 mmHg\right)÷\left(740mmHg\right) \\ {T}_2=287.62 K=\left(287.62-273\right)°C=14.62°C \end{gathered}$$