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The density of a gas can be determined using the ideal gas law, which relates pressure, temperature, volume, and number of moles. For density calculations, the equation is rearranged into the form:

d=PMRTd = \frac{PM}{RT}d=RTPM

Here, $d$ is density in g/L, $P$ is pressure in atm, $M$ is molar mass in g/mol, $R$ is the universal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹), and $T$ is absolute temperature in Kelvin.

For Argon (Ar), a noble gas, the molar mass is 39.9 g/mol. At the given conditions:

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  Pressure = 1.05 atm

  
  


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  Temperature = 30 °C = 303 K

  
  


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  Gas constant R = 0.0821 L·atm·K⁻¹·mol⁻¹

  
  

Substituting into the formula:

d=1.05×39.90.0821×303d = \frac{1.05 \times 39.9}{0.0821 \times 303}d=0.0821×3031.05×39.9 d=41.89524.8763≈1.68 g/Ld = \frac{41.895}{24.8763} \approx 1.68 \, g/Ld=24.876341.895≈1.68g/L

This result shows that under slightly above atmospheric pressure (1.05 atm) and room temperature (30 °C), the density of argon gas is 1.68 g/L.

This calculation is a practical application of Physical Chemistry principles, specifically gas laws. It demonstrates how gas density depends on both temperature and pressure. At higher pressures, density increases, while at higher temperatures, density decreases due to expansion of volume.

Argon, being a noble gas with very weak intermolecular forces, closely follows the ideal gas law, making this calculation accurate. Such gas law problems are fundamental for understanding real-life applications in chemical engineering, atmospheric chemistry, and laboratory experiments.

Thus, the correct density of Argon gas under the given conditions is 1.68 g/L (Option C).