Carbon dioxide in a gas cylinder of unchangeable volume is at a pressure of \(25.0 \mathrm{~atm}\) at \(25^{\circ} \mathrm{C}\). When placed in the sunlight on a hot summer day, the temperature increases to \(40^{\circ} \mathrm{C}\). What is the new pressure of the \(\mathrm{CO}_{2}\) ?

Short Answer

Expert verified
The new pressure of the \(\mathrm{CO}_{2}\) gas when the temperature increases to \(40^{\circ} \mathrm{C}\) is \(26.3 \mathrm{~atm}\).

Step by step solution

01

Write down the Gay-Lussac's Law equation

The equation for Gay-Lussac's Law is given as: \(P_1/T_1 = P_2/T_2\) Here, \(P_1\) is the initial pressure, \(T_1\) is the initial temperature, \(P_2\) is the final pressure and \(T_2\) is the final temperature.
02

Convert the temperatures to Kelvin

As we know, to work with gas laws, we need to have temperatures in Kelvin. Let's convert the given temperatures from Celsius to Kelvin: Initial temperature (\(T_1\)): \(25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{K}\) Final temperature (\(T_2\)): \(40^{\circ} \mathrm{C} + 273.15 = 313.15 \mathrm{K}\)
03

Substitute the given values into the equation

Now, we will substitute the given values into the Gay-Lussac's Law equation: \(\frac{25.0 \mathrm{~atm}}{298.15 \mathrm{K}} = \frac{P_2}{313.15 \mathrm{K}}\)
04

Solve for the final pressure (\(P_2\))

To find the new pressure of the \(\mathrm{CO}_{2}\) gas, rearrange and solve for \(P_2\): \(P_2 = \frac{25.0 \mathrm{~atm} \times 313.15 \mathrm{K}}{298.15 \mathrm{K}}\) Now, calculate the new pressure: \(P_2 = 26.3 \mathrm{~atm}\)
05

State the answer

The new pressure of the \(\mathrm{CO}_{2}\) gas when the temperature increases to \(40^{\circ} \mathrm{C}\) is \(26.3 \mathrm{~atm}\).

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