Answer :
Let's solve the problem step-by-step using the provided information:
1. Given Data:
- Volume of the lightbulb, [tex]\( V = 75.0 \)[/tex] milliliters
- Moles of neon gas, [tex]\( n = 7.16 \times 10^{-4} \)[/tex] moles
- Absolute pressure, [tex]\( P = 116.8 \)[/tex] kilopascals (kPa)
- Ideal gas constant, [tex]\( R = 8.314 \frac{ L \cdot kPa }{ mol \cdot K} \)[/tex]
- Standard atmospheric pressure, [tex]\( 1 \)[/tex] atm [tex]\( = 101.3 \)[/tex] kPa
2. Convert the volume from milliliters to liters:
[tex]\[ V = 75.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.0750 \text{ L} \][/tex]
3. Convert pressure from kilopascals to atmospheres:
[tex]\[ P_{atm} = \frac{116.8 \text{ kPa}}{101.3 \text{ kPa/atm}} \approx 1.153 \text{ atm} \][/tex]
4. Use the Ideal Gas Law [tex]\((PV = nRT)\)[/tex] to solve for the temperature [tex]\(T\)[/tex]:
[tex]\[ T = \frac{PV}{nR} \][/tex]
Substituting in the values, we get:
[tex]\[ T = \frac{(1.153 \text{ atm}) \times (0.0750 \text{ L})}{(7.16 \times 10^{-4} \text{ moles}) \times (0.0821 \frac{L \cdot atm}{mol \cdot K})} \][/tex]
5. Perform the calculation:
[tex]\[ T = \frac{0.086475 \text{ L} \cdot \text{ atm}}{0.000058828 \text{ L} \cdot \text{ atm} / \text{ K}} \][/tex]
[tex]\[ T \approx 14.527 \text{ K} \][/tex]
Expressing the final temperature to three significant figures:
The temperature of the lightbulb was [tex]\(\boxed{14.5} \text{ K}\)[/tex]
1. Given Data:
- Volume of the lightbulb, [tex]\( V = 75.0 \)[/tex] milliliters
- Moles of neon gas, [tex]\( n = 7.16 \times 10^{-4} \)[/tex] moles
- Absolute pressure, [tex]\( P = 116.8 \)[/tex] kilopascals (kPa)
- Ideal gas constant, [tex]\( R = 8.314 \frac{ L \cdot kPa }{ mol \cdot K} \)[/tex]
- Standard atmospheric pressure, [tex]\( 1 \)[/tex] atm [tex]\( = 101.3 \)[/tex] kPa
2. Convert the volume from milliliters to liters:
[tex]\[ V = 75.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.0750 \text{ L} \][/tex]
3. Convert pressure from kilopascals to atmospheres:
[tex]\[ P_{atm} = \frac{116.8 \text{ kPa}}{101.3 \text{ kPa/atm}} \approx 1.153 \text{ atm} \][/tex]
4. Use the Ideal Gas Law [tex]\((PV = nRT)\)[/tex] to solve for the temperature [tex]\(T\)[/tex]:
[tex]\[ T = \frac{PV}{nR} \][/tex]
Substituting in the values, we get:
[tex]\[ T = \frac{(1.153 \text{ atm}) \times (0.0750 \text{ L})}{(7.16 \times 10^{-4} \text{ moles}) \times (0.0821 \frac{L \cdot atm}{mol \cdot K})} \][/tex]
5. Perform the calculation:
[tex]\[ T = \frac{0.086475 \text{ L} \cdot \text{ atm}}{0.000058828 \text{ L} \cdot \text{ atm} / \text{ K}} \][/tex]
[tex]\[ T \approx 14.527 \text{ K} \][/tex]
Expressing the final temperature to three significant figures:
The temperature of the lightbulb was [tex]\(\boxed{14.5} \text{ K}\)[/tex]