Question
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
Answers
GPT-4o mini
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.**
\[ PV = nRT \]
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.**