Question

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

a. A sample of xenon gas at $25°C$and $740mmHg$is cooled to give a pressure of $620mmHg$.

b. A tank of argon gas with a pressure of$0950atm$at$-18°C$is heated to give a pressure of$1250Torr.$

Short answer

a. The final temperature of xenon gas is$-23.32°C$.

b. The final temperature of argon gas is role="math" localid="1652606265951" $167.2C°$.

Step-by-step solution

Part (a) step 1: Given Information 

We need to find the final temperature of the xenon 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 $T_1=25°C=\left(25+273\right)K=298K$

initial pressure $P_1=740mmHg$

the final pressure $P_2=620mmHg$

Now, calculating the final temperature $T_2$:

localid="1652607707341" $$\begin{gathered} T_2=\left(T_1\right)\times \left(P_2\right)÷\left(P_1\right) \\ {T}_2=\left(298K\right)\times \left(620mmHg\right)÷\left(740mmHg\right) \\ {T}_2=249.67K=\left(249.67-273\right)°C=-23.32°C \end{gathered}$$

Part (a) step 1: Given Information

We need to find the final temperature of argon 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$T_2=-18°=\left(-18+273\right)K=255K$

initial pressure$P_1=095atm$

the final pressure$1250torr=1.64atm$

Now, calculating the final temperature$T_2$

localid="1652607722006" $$\begin{gathered} T_2=\left(T_1\right)\times \left(P_2\right)÷\left(P_1\right) \\ {T}_2=\left(255K\right)\times \left(1.64atm\right)÷\left(0.95atm\right) \\ {T}_2=440.21K=\left(440.21-273\right)°C=167.2°C \end{gathered}$$