one mole of an ideal gas expanded isothermally and reversibly at temperature of 300k from 10 to 1 atmospheric (atm). Calculate the work.
Well, let's calculate that work one joke at a time! Why did the ideal gas go on a diet? Because it wanted to lose some pressure, of course!
Now, to calculate the work done by the gas during an isothermal expansion, we can use the formula:
Work = nRT ln(Vf/Vi)
Where:
- n is the number of moles of gas (in this case, n = 1)
- R is the ideal gas constant (R = 0.0821 L·atm/mol·K)
- T is the temperature in Kelvin (T = 300 K)
- Vf is the final volume (Vf = 1 atm)
- Vi is the initial volume (Vi = 10 atm)
Now, let's put these values in and let the math do the funny stuff!
Work = (1 mol) * (0.0821 L·atm/mol·K) * 300 K * ln(1/10)
Work = -221.8 L·atm
So, the work done by the ideal gas is approximately -221.8 L·atm. And by the way, negative work indicates that the gas is doing work on the surroundings.
Remember, laughter is always the ideal gas!
To calculate the work done during an isothermal and reversible expansion of an ideal gas, you can use the following equation:
Work (W) = -nRT ln(V2/V1)
Where:
- n is the number of moles of the gas
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- V2 is the final volume
- V1 is the initial volume
In this case, you have n = 1 mole, T = 300 K, V1 = 10 atm, and V2 = 1 atm.
Now, let's calculate the work:
Work (W) = -1 mole * (8.314 J/mol·K) * 300 K * ln(1/10)
Work (W) = -1 * 8.314 J/K * 300 K * ln(0.1)
Work (W) ≈ -1 * 8.314 J/K * 300 K * (-2.3026)
Work (W) ≈ 4859.4 J
Therefore, the work done during this isothermal and reversible expansion is approximately 4859.4 J.
To calculate the work done during the isothermal expansion of an ideal gas, we can use the equation:
Work = -nRT * ln(V2/V1)
Where:
- Work is the work done by or on the gas during the process
- n is the number of moles of the gas
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- V1 is the initial volume of the gas
- V2 is the final volume of the gas
Given:
- n = 1 mole
- R = 8.314 J/mol·K
- T = 300 K
- V1 = 10 atm
- V2 = 1 atm
Now, let's substitute these values into the equation to calculate the work:
Work = - (1 mole) * (8.314 J/mol·K) * 300 K * ln(1/10)
First, let's calculate the natural logarithm:
ln(1/10) ≈ -2.3026
Now, substitute this back into the previous equation:
Work = - (1 mole) * (8.314 J/mol·K) * 300 K * (-2.3026)
Work ≈ 5509.412 J
Therefore, the work done during the isothermal expansion of the ideal gas is approximately 5509.412 J.