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A sample of gas has a volume of 80 o mL at -23.0^circ C and 300 Torr. What would the volume of the gas be at 227.0^circ C and 600 Torr? a 500 mL b 800 mL c 200 mL d 100 mL

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A sample of gas has a volume of 80 o mL at -23.0^circ C and 300 Torr. What would the volume of the gas be at 227.0^circ C and 600 Torr? a 500 mL b 800 mL c 200 mL d 100 mL

A sample of gas has a volume of 80 o mL at -23.0^circ C and 300 Torr. What would the volume of the gas be at 227.0^circ C and 600 Torr? a 500 mL b 800 mL c 200 mL d 100 mL

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NigellaProfessional · Tutor for 6 years

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### b \( 800 \mathrm{~mL} \)

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## Step 1: Identify the ideal gas law to use<br />### The combined gas law, which relates initial and final states of pressure, volume, and temperature, is: <br />\[<br />\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}<br />\]<br />## Step 2: Convert temperatures to Kelvin<br />### Initial temperature \( T_1 = -23.0^{\circ} \mathrm{C} + 273.15 = 250.15 \mathrm{K} \)<br />### Final temperature \( T_2 = 227.0^{\circ} \mathrm{C} + 273.15 = 500.15 \mathrm{K} \)<br />## Step 3: Plug in given values to the combined gas law<br />### Given: \( P_1 = 300 \, \text{Torr} \), \( V_1 = 800 \, \text{mL} \), \( P_2 = 600 \, \text{Torr} \), find \( V_2 \)<br />\[<br />\frac{300 \, \text{Torr} \cdot 800 \, \text{mL}}{250.15 \, \text{K}} = \frac{600 \, \text{Torr} \cdot V_2}{500.15 \, \text{K}}<br />\]<br />## Step 4: Simplify and solve for \( V_2 \)<br />\[<br />V_2 = \frac{300 \cdot 800 \cdot 500.15}{250.15 \cdot 600} \, \text{mL}<br />\]<br />\[<br />V_2 \approx 800 \, \text{mL}<br />\]
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