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How many moles of a gas sample are in a 10.0 L container at 298 K and 203 kPa? Use the ideal gas law [tex]\( PV = nRT \)[/tex] with [tex]\( R = 8.31 \, \text{L} \cdot \text{kPa} / \text{mol} \cdot \text{K} \)[/tex].

A. 0.60 mole
B. 0.82 mole
C. 1.22 moles
D. 1.66 moles


Sagot :

To find out how many moles of a gas sample are in a 10.0 L container at 298 K and 203 kPa, we will use the ideal gas law:

[tex]\[ PV = nRT \][/tex]

Here, [tex]\( P \)[/tex] is the pressure, [tex]\( V \)[/tex] is the volume, [tex]\( n \)[/tex] is the number of moles, [tex]\( R \)[/tex] is the gas constant, and [tex]\( T \)[/tex] is the temperature. We are given the following values:

- [tex]\( P = 203 \)[/tex] kPa
- [tex]\( V = 10.0 \)[/tex] L
- [tex]\( T = 298 \)[/tex] K
- [tex]\( R = 8.31 \)[/tex] L·kPa/mol·K

We need to solve for [tex]\( n \)[/tex], the number of moles. Rearranging the ideal gas law to solve for [tex]\( n \)[/tex]:

[tex]\[ n = \frac{PV}{RT} \][/tex]

Substituting the given values into the equation:

[tex]\[ n = \frac{(203 \, \text{kPa})(10.0 \, \text{L})}{(8.31 \, \text{L·kPa/mol·K})(298 \, \text{K})} \][/tex]

Upon calculating, we get:

[tex]\[ n = \frac{2030 \, \text{kPa·L}}{2476.38 \, \text{L·kPa/mol·K}} \][/tex]

Approximately:

[tex]\[ n \approx 0.82 \, \text{moles} \][/tex]

Thus, the number of moles of the gas sample in the container is approximately [tex]\(0.82 \, \text{moles}\)[/tex].

Therefore, the correct answer is:

0.82 mole