Question

A sample of \(\mathrm{CO}_{2}\) gas at \(22{ }^{\circ} \mathrm{C}\) and 1.00 atm has a volume of \(2.00 \mathrm{~L} .\) Determine the ratio of the original volume to the final volume when (a) the pressure and amount of gas remain unchanged and the Celsius temperature is doubled. (b) the pressure and amount of gas remain unchanged and the Kelvin temperature is doubled.

Short answer

Answer: (a) The ratio of the original volume to the final volume is approximately 1.07 when the Celsius temperature is doubled. (b) The ratio of the original volume to the final volume is 2 when the Kelvin temperature is doubled.

Step-by-step solution

Label initial and final conditions

In both cases, we'll call the initial volume \(V_1\), the initial temperature \(T_1\), the final volume \(V_2\), and the final temperature \(T_2\). We begin with: \(V_1 = 2.00 \mathrm{~L}, T_1 = 22{ }^{\circ} \mathrm{C}\) and 1.00 atm pressure.

Convert initial temperature to Kelvin

To convert the initial temperature from Celsius to Kelvin, we use the formula: \(T_K = T_C + 273.15\). So, our initial temperature is: \(T_1 = 22 + 273.15 = 295.15\mathrm{~K}\). (a)

Step 3a: Double the Celsius temperature for part (a)

To find the final temperature for part (a), we need to double the initial Celsius temperature and then convert it to Kelvin: \(T_{2a} = 2 * 22 = 44{ }^{\circ} \mathrm{C}\) and \(T_{2a} = 44 + 273.15 = 317.15\mathrm{~K}\).

Step 4a: Find the volume ratio for part (a)

Using the proportionality \(V \propto T\), we can write the volume ratio: \((V_2/V_1) = (T_2/T_1)\). For part (a), this becomes: \((V_2/V_1) = (T_{2a}/T_1) = (317.15/295.15)\). The volume ratio for part (a) is approximately \(1.07\). (b)

Step 3b: Double the Kelvin temperature for part (b)

To find the final temperature for part (b), we need to double the initial Kelvin temperature: \(T_{2b} = 2 * 295.15 = 590.3\mathrm{~K}\)

Step 4b: Find the volume ratio for part (b)

Again using the proportionality, we can write the volume ratio: \((V_2/V_1) = (T_2/T_1)\). For part (b), this becomes: \((V_2/V_1) = (T_{2b}/T_1) = (590.3/295.15)\). The volume ratio for part (b) is \(2\). In conclusion, (a) the ratio of the original volume to the final volume is approximately \(1.07\) when the Celsius temperature is doubled, and (b) the ratio of the original volume to the final volume is \(2\) when the Kelvin temperature is doubled.