What is the temperature, in °C, of a 0.4 mole sample of an ideal gas contained in a volume of 3,608 mL at 1,769 torr? Round your answer to the nearest degree.



Answer :

To determine the temperature of a 0.4 mole sample of an ideal gas contained in a volume of 3,608 mL at 1,769 torr, follow these steps:

1. Convert the volume from milliliters to liters:
- The given volume is [tex]\( 3,608 \)[/tex] mL.
- To convert milliliters to liters:
[tex]\[ \text{Volume in Liters} = \frac{3,608 \text{ mL}}{1,000} = 3.608 \text{ L} \][/tex]

2. Use the Ideal Gas Law (PV = nRT) to find the temperature in Kelvin:
- Given values:
- Pressure [tex]\( P = 1,769 \)[/tex] torr
- Volume [tex]\( V = 3.608 \)[/tex] L
- Moles [tex]\( n = 0.4 \)[/tex] moles
- The gas constant [tex]\( R = 62.3637 \)[/tex] Torr·L / (mol·K)
- Rearrange the Ideal Gas Law to solve for temperature [tex]\( T \)[/tex]:
[tex]\[ T = \frac{PV}{nR} \][/tex]
- Substitute the given values:
[tex]\[ T = \frac{1,769 \text{ torr} \times 3.608 \text{ L}}{0.4 \text{ moles} \times 62.3637 \text{ Torr·L / (mol·K)}} \][/tex]
- Calculate the temperature [tex]\( T \)[/tex] in Kelvin:
[tex]\[ T = \frac{1,769 \times 3.608}{0.4 \times 62.3637} = 255.8601 \text { K} \][/tex]

3. Convert the temperature from Kelvin to Celsius:
- The conversion formula is [tex]\( T_C = T_K - 273.15 \)[/tex]:
[tex]\[ T_C = 255.8601 \text{ K} - 273.15 = -17.2899 \text { °C} \][/tex]

4. Round the temperature to the nearest degree:
- Rounded temperature:
[tex]\[ -17.2899 \text{ °C} \approx -17 \text{ °C} \][/tex]

Thus, the temperature of the ideal gas sample is -17°C.