10 mol of an ideal gas is expanded reversibly isothermally from 1 L to 10 L at 100 K. The work done during the process is:–1.91 kJ–0.95 kJ–19.1 kJ–38.2 kJ
Question
10 mol of an ideal gas is expanded reversibly isothermally from 1 L to 10 L at 100 K. The work done during the process is:–1.91 kJ–0.95 kJ–19.1 kJ–38.2 kJ
Solution
To calculate the work done during the isothermal expansion of the ideal gas, we can use the formula:
Work = -nRT ln(V2/V1)
Where:
- n is the number of moles of the gas (given as 10 mol)
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin (given as 100 K)
- V1 is the initial volume (given as 1 L)
- V2 is the final volume (given as 10 L)
Plugging in the values, we have:
Work = -10 mol * 8.314 J/mol·K * 100 K * ln(10 L / 1 L)
Simplifying the equation, we get:
Work = -10 * 8.314 * 100 * ln(10)
Using a calculator, we find that ln(10) is approximately 2.3026. Substituting this value, we have:
Work = -10 * 8.314 * 100 * 2.3026
Calculating this expression, we find that the work done during the process is approximately -1910 J, which is equal to -1.91 kJ.
Therefore, the correct answer is –1.91 kJ.
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