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0.650 moles of O₂(g) are added to 1.00 L flask and the internal pressure is measured at 20.00 atm. What is the temperature of the gas under these conditions (in *C)?___ °C ?

Respuesta :

Answer:

T = 101.96 degrees

Explanation:

We are given the following

Number of moles = 0.650 mol

Volume = 1.00 L

Pressure = 20.00 atm

We are going to use the ideal gas low which is given by:

pV = nRT

Therefore:

[tex]T\text{ = }\frac{pV}{nR}[/tex]

We know R = 0.08206 L.atm/Kmol

[tex]\begin{gathered} T\text{ = }\frac{20\ast1}{0.65\ast0.08206} \\ T\text{ = 374.960 K} \end{gathered}[/tex]

T(in degrees) = 374.96 - 273

= 101.96 degrees

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