To determine the temperature of the gas sample, we can use the Ideal Gas Law, which is expressed as:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure of the gas,
- [tex]\( V \)[/tex] is the volume of the gas,
- [tex]\( n \)[/tex] is the number of moles of the gas,
- [tex]\( R \)[/tex] is the Universal Gas Constant, and
- [tex]\( T \)[/tex] is the temperature in Kelvin.
Let's analyze the given values:
- The number of moles ([tex]\( n \)[/tex]) of the gas is 0.399 moles.
- The volume ([tex]\( V \)[/tex]) of the gas is 3,980 mL.
- The pressure ([tex]\( P \)[/tex]) is 1,828 torr.
- The Universal Gas Constant ([tex]\( R \)[/tex]) in units of L·torr/(K·mol) is 62.364 L·torr/(K·mol).
Before using these values in the Ideal Gas Law equation, we need to ensure that the units are consistent. Specifically, the volume should be in liters (L) rather than milliliters (mL). To convert the volume from milliliters to liters, we divide by 1,000, since 1 L = 1,000 mL:
[tex]\[ V = \frac{3980 \text{ mL}}{1000 \text{ mL/L}} = 3.98 \text{ L} \][/tex]
Now we have:
- [tex]\( n = 0.399 \)[/tex] moles,
- [tex]\( V = 3.98 \)[/tex] L,
- [tex]\( P = 1,828 \)[/tex] torr,
- [tex]\( R = 62.364 \)[/tex] L·torr/(K·mol).
We can now rearrange the Ideal Gas Law to solve for the temperature [tex]\( T \)[/tex]:
[tex]\[ T = \frac{PV}{nR} \][/tex]
Substituting in the given values, we get:
[tex]\[ T = \frac{(1828 \text{ torr})(3.98 \text{ L})}{(0.399 \text{ mol})(62.364 \text{ L·torr/(K·mol)})} \][/tex]
After performing the calculation and rounding to the nearest degree, we find that the temperature [tex]\( T \)[/tex] is:
[tex]\[ T \approx 292 \text{ K} \][/tex]
Therefore, the temperature of the 0.399 mole sample of an ideal gas contained in a volume of 3,980 mL at 1,828 torr, rounded to the nearest degree, is 292 K.
