Eight grams of helium (molecular mass = 4.0026 u) expand isothermally at 340 K and does 9550 J of work. Assuming that helium is an ideal gas, determine the ratio of the final volume of the gas to the initial volume.
The Correct Answer and Explanation is :
We can use the thermodynamic equation for an isothermal expansion of an ideal gas to solve this problem:
[
W = nRT \ln \left( \frac{V_f}{V_i} \right)
]
Where:
- ( W ) is the work done by the gas (9550 J),
- ( n ) is the number of moles of gas,
- ( R ) is the ideal gas constant ((8.314 \, \text{J/mol·K})),
- ( T ) is the temperature (340 K),
- ( V_f ) is the final volume,
- ( V_i ) is the initial volume.
Step 1: Determine the number of moles ((n))
The number of moles of helium can be calculated using its mass and molar mass:
[
n = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \, \text{g}}{4.0026 \, \text{g/mol}} \approx 2.000 \, \text{mol}
]
Step 2: Rearrange the work equation to solve for the volume ratio
[
W = nRT \ln \left( \frac{V_f}{V_i} \right)
]
Solving for the volume ratio (\frac{V_f}{V_i}):
[
\ln \left( \frac{V_f}{V_i} \right) = \frac{W}{nRT}
]
Substitute the known values:
[
\ln \left( \frac{V_f}{V_i} \right) = \frac{9550 \, \text{J}}{2.000 \, \text{mol} \times 8.314 \, \text{J/mol·K} \times 340 \, \text{K}}
]
[
\ln \left( \frac{V_f}{V_i} \right) = \frac{9550}{5623.56} \approx 1.698
]
Step 3: Solve for the ratio ( \frac{V_f}{V_i} )
[
\frac{V_f}{V_i} = e^{1.698} \approx 5.46
]
Final Answer:
The ratio of the final volume to the initial volume is approximately 5.46.
Explanation:
In an isothermal expansion, the temperature remains constant, and the gas does work on its surroundings. The equation used here is derived from the first law of thermodynamics, where the internal energy change is zero for an ideal gas during an isothermal process. The work done by the gas depends on the initial and final volumes, and the ratio of these volumes can be found using the relationship derived from the ideal gas law. The positive work indicates that the gas expanded, doing work by pushing against its surroundings.