Question

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

Answer 1

The answer is "
`2.73 * 10^3 \ K` ".

Step-by-step explanation:

Please find the complete question in the attachment.

The Formula for Ideal gas:

`\to PV = nRT \\\\`

`= ( (m)/(M))RT`

`\to Density\ \rho = (m)/(V) = (PM)/(RT)`

`\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`

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