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6. What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0^circ C to 30.0^circ C and a pressure change from 1 atm to 8.6 atm? square

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6. What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change
from 22.0^circ C to 30.0^circ C and a pressure change from 1 atm to 8.6 atm?
square

6. What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0^circ C to 30.0^circ C and a pressure change from 1 atm to 8.6 atm? square

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AvaVeteran · Tutor for 12 years

Answer

### 48.76 \, \mathrm{mL}

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## Step 1: Identify the Gas Law<br />### Use the combined gas law: \( PV/T = \text{constant} \).<br />## Step 2: Convert Temperatures to Kelvin<br />### \( 22.0^{\circ} \mathrm{C} = 295.15 \, \mathrm{K} \)<br />### \( 30.0^{\circ} \mathrm{C} = 303.15 \, \mathrm{K} \)<br />## Step 3: Use the Combined Gas Law<br />### Substitute values: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).<br />### \( \frac{1 \, \mathrm{atm} \times 400.0 \, \mathrm{mL}}{295.15 \, \mathrm{K}} = \frac{8.6 \, \mathrm{atm} \times V_2}{303.15 \, \mathrm{K}} \)<br />## Step 4: Solve for \( V_2 \)<br />### Rearrange and substitute to find \( V_2 \).<br />### \( V_2 = \frac{1 \times 400.0 \times 303.15}{8.6 \times 295.15} \approx 48.76 \, \mathrm{mL} \)
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