Which equation is set up correctly to determine the volume of a 3.2 mole sample of oxygen gas at 50°C and 101.325 kPa?

A V=(3.2 mol)(8.314 L⋅kPa/K⋅mol)(323 K)101.325 kPa

B V=(3.2 mol)(8.314 L⋅kPa/K⋅mol)(50∘C)1 atm

C V=(8.314 L⋅kPa/K⋅mol)(50∘C)(3.2 mol)(101.325 kPa)

D V=(3.2 mol)(8.314 L⋅kPa/K⋅mol)(1 atm)(323 K)

n = 3.2 moles, T = 50 + 273 = 323 K, P = 101.325 kPa, R = 8.314 L.kPa/K.mol

PV = nRT

V = nRT / P substituting.

V = (3.2 mole)(8.314 L.kPa/K.mol )(323 K) / (101.325 kPa)

That is the answer, but it is not among the options you provided. Check your options properly.

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dexteright02 dexteright02 · Answer 2 · 2018-10-24T23:54:07+02:00

Hello!

Which equation is set up correctly to determine the volume of a 3.2 mole sample of oxygen gas at 50°C and 101.325 kPa?

We have the following data:

v (volume) = ? (in L)

n (number of mols) = 3.2 mol

T (temperature) = 50 ºC

First let's convert the temperature on the Kelvin scale, let's see:

TK = TºC + 273

TK = 50 + 273

TK = 323

P (pressure) = 101.325 kPa

R (gas constant) = 8.314 L . kPa/K.mol

We apply the data above to the Clapeyron equation (gas equation), let's see:

[tex]P*V = n*R*T[/tex]

[tex]101.325\:kPa*V = \left(3.2\:mol)\right*\left(8.314\:L.kPa/K.mol)\right*323\:K[/tex]

[tex]\boxed{\boxed{V = \dfrac{\left(3.2\:mol)\right*\left(8.314\:L.kPa/K.mol)\right*323\:K}{101.325\:kPa} }}\end{array}}\qquad\checkmark[/tex]

Answer:

(A) V=(3.2 mol)(8.314 L⋅kPa/K⋅mol)(323 K) / 101.325 kPa

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I Hope this helps, greetings ... Dexteright02! =)