Question

Question:The pressure of a poisonous gas inside a sealed container is 1.47 atm at 20°C. If the barometric pressure is 0.96 atm, to what temperature (in degrees Celsius) must the container and its contents be cooled so that the container can be opened with no risk for gas spurting out?

Short answer

Answer

The container and its contents must be cooled to -81.8⁰C so that we can open the container without the risk of gas spurting out.

Step-by-step solution

Gay-Lussac law

Gay-Lussac’s law states that when the volume is constant, the pressure of an ideal gas is directly proportional to the absolute temperature.

Step 2: Calculation of temperature

The pressure of the gas must be less than 0.96 atm to prevent the gas spurting from the container.

It is given that the initial pressure, ${\text{P}}_1\text{=}1.47\text{atm}$

The initial temperature, ${\text{T}}_1=20°\text{C = 273 + 20 = 293}\text{K}$

The final pressure,${\text{P}}_2=0.96\text{atm}$ ${\text{P}}_2=0.96\text{atm}$

The final temperature, ${\text{T}}_2=?$

Substitute these values in the equation given below:

$$\begin{gathered} \frac{{\text{P}}_{\text{1}}}{{\text{T}}_{\text{1}}}=\frac{{\text{P}}_{\text{2}}}{{\text{T}}_{\text{2}}} \\ \frac{1.47\text{atm}}{\text{293}\text{K}}=\frac{0.96\text{atm}}{{\text{T}}_{\text{2}}} \\ {\text{T}}_{\text{2}}=\frac{0.96\text{atm}\times \text{293}\text{K}}{1.47\text{atm}} \\ =191.35\text{K} \end{gathered}$$

The temperature in kelvin is 191.35 K.

Conversion of kelvin to degree celsius

The temperature in kelvin is converted to degree Celsius by the following method:

$$\begin{gathered} {}^{\circ }\text{C = 273 - K} \\ °\text{C = K - 273} \\ =191.35-273 \\ =-81.8°\text{C} \end{gathered}$$

Hence, the temperature in degree Celsius is$-81.8°\text{C}$ .

The container and its contents must be cooled to $-81.8°\text{C}$ so that we can open the container without the risk of gas spurting out.