Question

a sample of neon gas at 50.°c and a volume of 2.5 l is cooled to 25°c. what is the new volume? use the formula: \\(\frac{v_1}{t_1} = \frac{v_2}{t_2}\\) \\(\circ\\) 2.3 l \\(\circ\\) 2.7 l \\(\circ\\) 5.0 l \\(\circ\\) 32.3 l done

Explanation:

Step1: Convert temperatures to Kelvin

To use the gas law formula, temperatures must be in Kelvin. The conversion formula is \( T(K) = T(^\circ C) + 273.15 \).
For \( T_1 \): \( T_1 = 50. + 273.15 = 323.15 \, K \)
For \( T_2 \): \( T_2 = 25 + 273.15 = 298.15 \, K \)

Step2: Identify known values

We know:
\( V_1 = 2.5 \, L \), \( T_1 = 323.15 \, K \), \( T_2 = 298.15 \, K \)
We need to find \( V_2 \). The formula is \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \).

Step3: Rearrange the formula to solve for \( V_2 \)

Multiply both sides by \( T_2 \):
\( V_2 = V_1 \times \frac{T_2}{T_1} \)

Step4: Substitute the values into the formula

Substitute \( V_1 = 2.5 \, L \), \( T_1 = 323.15 \, K \), \( T_2 = 298.15 \, K \):
\( V_2 = 2.5 \, L \times \frac{298.15}{323.15} \)

Step5: Calculate the result

First, compute \( \frac{298.15}{323.15} \approx 0.9226 \).
Then, \( V_2 \approx 2.5 \times 0.9226 \approx 2.3 \, L \)

Answer:

2.3 L