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What is the volume of a gas if 0.182 moles of the gas is at 1.99 atm and 83.4oC?

Sagot :

[tex]\text{Given that,}\\\\\text{Pressure,}~ P = 1.99~ \text{atm}\\ \\\text{No. of moles,}~ n = 0.182~ \text{mol.}\\\\\text{Temperature,}~ T = 83.4^{\circ}~ C = 356.4~K\\\\ \text{Molar gas constant,}~ R = 0.082~ \text{L}~ \text{atm}~ \text{mol}^{-1}~ \text{K}^{-1}\\\\\text{Volume,}~ V =?\\\\\text{We know that,}~\\\\~~~~~~~PV =nRT\\\\\implies V = \dfrac{nRT}{P}\\\\\\~~~~~~~~~~=\dfrac{0.182 \times0.082 \times 356.4 }{1.99}\\\\\\~~~~~~~~~~\approx 2.68~ \text{L.}\\\\\\[/tex]

[tex]\text{Hence the volume of the gas is 2.68 L.}[/tex]

Answer:

19923

Explanation:

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