To solve this problem, we’ll apply Charles’s Law, which describes how the volume of a gas changes with temperature, assuming that pressure and the amount of gas remain constant.
Charles’s Law is given by the formula:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where:
\( V_1 \) is the initial volume of the gas,
\( T_1 \) is the initial temperature (in Kelvin) of the gas,
\( V_2 \) is the final volume of the gas,
\( T_2 \) is the final temperature (in Kelvin) of the gas.
Now, let's follow these steps:
**Step 1: Convert temperatures from Celsius to Kelvin.**
Since temperature must be measured on an absolute scale, we convert Celsius to Kelvin using the conversion:
\[ T(Kelvin) = T(Celsius) + 273.15 \]
For the initial temperature, \( T_1 \) (86.8°C):
\[ T_1 = 86.8°C + 273.15 = 359.95 K \]
For the final temperature, \( T_2 \) (31.9°C):
\[ T_2 = 31.9°C + 273.15 = 305.05 K \]
**Step 2: Insert the known values into Charles's Law and solve for the final volume \( V_2 \).**
Initial volume \( V_1 = 287.6 \) mL. We have:
\[ \frac{287.6 \text{ mL}}{359.95 K} = \frac{V_2}{305.05 K} \]
Now, cross multiply to solve for \( V_2 \):
\[ V_2 = \frac{287.6 \text{ mL} \times 305.05 K}{359.95 K} \]
**Step 3: Perform the calculation.**
\[ V_2 = \frac{287.6 \text{ mL} \times 305.05 K}{359.95 K} \]
\[ V_2 ≈ \frac{87719.28 \text{ mL} \cdot K}{359.95 K} \]
\[ V_2 ≈ 243.69 \text{ mL} \]
**Step 4: State the final answer.**
The volume that the sample of Nitrogen gas will occupy at 31.9°C is approximately 243.69 mL.
Note that our calculation might differ slightly from the exact answer due to rounding the temperature conversions and the final calculation.
