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 role="math" localid="1652606588066" $250Torr$at $0°C$is heated to give a pressure of role="math" localid="1652606595080" $1500Torr$.

b. A sample of air atrole="math" localid="1652606709077" $40°C$and$740mmHg$is cooled to give a pressure of$680mmHg.$

Short answer

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

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

Step-by-step solution

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$P_1=250Torr$

the final pressure$P_2=1500Torr$

Now, calculating the final temperature $T_2$:

localid="1652607783216" role="math" $$\begin{gathered} T_2=\left(T_1\right)\times \left(P_2\right)÷\left(P_1\right) \\ {T}_2=\left(273K\right)\times \left(1500mmHg\right)÷\left(250mmHg\right) \\ {T}_2=1638K=\left(1638-273\right)°C=1365°C \end{gathered}$$

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 pressurelocalid="1652607317434" $P_1=740mmHg$

the final pressure$P_2=680mmHg$

Now, calculating the final temperature $T_2$:

localid="1652607796504" $$\begin{gathered} T_2=\left(T_1\right)\times \left(P_2\right)÷\left(P_1\right) \\ {T}_2=\left(313K\right)\times \left(680mmHg\right)÷\left(740mmHg\right) \\ {T}_2=287.62K=\left(287.62-273\right)°C=14.62°C \end{gathered}$$