The molar volume of a gas increases by a factor of 2. Which parameter change could have caused the shift in molar volume?(1 point) Responses the temperature increasing by a factor of 2 the temperature increasing by a factor of 2 the pressure increasing by a factor of 2 the pressure increasing by a factor of 2 the temperature decreasing by a factor of 0.25 the temperature decreasing by a factor of 0.25 the pressure decreasing by a factor of 0.25

1 answer

To determine which parameter change could have caused the molar volume of a gas to increase by a factor of 2, we can use the Ideal Gas Law, which is given by:

\[ PV = nRT \]

Where:

  • \( P \) = pressure
  • \( V \) = volume (molar volume when considering one mole of gas)
  • \( n \) = number of moles (which we will consider as 1 for simplicity)
  • \( R \) = universal gas constant
  • \( T \) = temperature (in Kelvin)

If the molar volume doubles (\( V \) increases by a factor of 2), we can look at the effects of changing temperature \( (T) \) and pressure \( (P) \).

  1. Temperature Increasing by a Factor of 2:

    • Doubling the temperature while keeping pressure constant would increase volume as \( V \propto T \), so this could theoretically increase the molar volume, but not necessarily by exactly a factor of 2 without knowing the pressure.
  2. Pressure Increasing by a Factor of 2:

    • Increasing pressure would decrease the volume (molar volume), so this cannot cause an increase in molar volume.
  3. Temperature Decreasing by a Factor of 0.25:

    • Reducing temperature would lead to a decrease in volume, thus not causing an increase.
  4. Pressure Decreasing by a Factor of 0.25:

    • Decreasing pressure would increase the volume (molar volume), and if pressure decreases to a quarter of its original value while temperature remains constant, this would increase volume significantly, resulting in a potential increase by a factor of 2 or more depending on the initial conditions.

Therefore, the correct answer is: the pressure decreasing by a factor of 0.25.