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What is the volume of 0.200 mol of an ideal gas at [tex]200 \, \text{kPa}[/tex] and [tex]400 \, \text{K}[/tex]?

Use [tex]PV = nRT[/tex] and [tex]R = 8.314 \frac{L \cdot kPa}{mol \cdot K}[/tex].

A. [tex]0.83 \, L[/tex]
B. [tex]3.33 \, L[/tex]
C. [tex]5.60 \, L[/tex]
D. [tex]20.8 \, L[/tex]

Sagot :

GinnyAnswer
To solve for the volume [tex]\( V \)[/tex] of an ideal gas using the ideal gas law [tex]\( PV = nRT \)[/tex], we need to rearrange the equation to solve for [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]

Let's plug in the given values:
- [tex]\( P \)[/tex] (Pressure) = 200 kPa
- [tex]\( n \)[/tex] (Number of moles) = 0.200 mol
- [tex]\( R \)[/tex] (Ideal gas constant) = 8.314 \frac{L \cdot kPa}{mol \cdot K}
- [tex]\( T \)[/tex] (Temperature) = 400 K

Now, substitute these values into the equation:
[tex]\[ V = \frac{0.200 \, \text{mol} \times 8.314 \, \frac{L \cdot kPa}{mol \cdot K} \times 400 \, K}{200 \, kPa} \][/tex]

Perform the multiplication and division:
[tex]\[ V = \frac{0.200 \times 8.314 \times 400}{200} \][/tex]

[tex]\[ V = \frac{665.12}{200} \][/tex]

[tex]\[ V = 3.3256 \, L \][/tex]

Therefore, the volume of the gas is approximately:
[tex]\[ V = 3.33 \, L \][/tex]

So, the correct answer is:
[tex]\[ \boxed{3.33 \, L} \][/tex]
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