Question
At 273 $\mathrm{K}$ and $1.00 \times 10^{-2}$ atm, the density of a gas is $1.24 \times$ $10^{-5} \mathrm{g} / \mathrm{cm}^{3} .(\mathrm{a})$ Find $v_{\mathrm{rms}}$ for the gas molecules. (b) Find the molarmass of the gas and (c) identify the gas. See Table $19-1 .$
Step 1
00 \times 10^{-2} \, atm = 1.01 \times 10^{3} \, Pa$ (using $1 \, atm = 1.01 \times 10^{5} \, Pa$), and the density $\rho = 1.24 \times 10^{-5} \, g/cm^{3} = 1.24 \times 10^{-2} \, kg/m^{3}$ (converting to SI units). Show more…
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At a temperature of 273 K and a pressure of 1.00 × 10^(-2) atm, the density of a monatomic gas is 1.24 × 10^(-5) g/cm^3. Also, we know 1 atm = 1.01 × 10^5 Pa, R = 8.31 J/mol·K, and k = 1.38 × 10^(-23) J/K. (a) Find vrms for the gas particles. (b) Find the molar mass of the gas.
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