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At what temperature does 1.00 atm of He gas have the same density as 1.00 atm?

Sagot :

Answer:

The answer is "[tex]2.73 \times 10^3 \ K[/tex] ".

Explanation:

Please find the complete question in the attachment.

The Formula for Ideal gas:

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

          [tex]= ( \frac{m}{M})RT[/tex]

[tex]\to Density\ \rho = \frac{m}{V} = \frac{PM}{RT}[/tex]

[tex]\to P= pressure\\\\\to V = volume\\\\ \to n = moles\ of \ gas \\\\\to R = molar \ gas \ constant\\\\ \to T = temperature\\\\ \to m = mass \\\\ \to M = molar \ mass[/tex]

[tex]\to P(Ar) = P(He) = 1.00 atm\\\\\to T(Ar) = ?\\\\ T(He) = 273.2 \ K\\\\\to M(Ar) = 39.948 \ \frac{g}{mol}\\\\ \to M(He) = 4.0026 \ \frac{g}{mol}\\\\\to \rho(Ar) = \rho(He)\\\\\bold{Formula: } \\\\ \to \frac{P(Ar)M(Ar)}{RT(Ar)} = \frac{P(He)M(He)}{RT(He)}\\\\\to \frac{1.00 \times 39.948}{(0.08206 \times T(Ar))} = \frac{1.00 \times 4.0026}{(0.08206 \times 273.2)}\\\\ \to T(Ar) = 2.73 \times 10^3 \ K[/tex]

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