A sample of gas in a cylinder of volume 3.78 L
at 318 K and 3.3 atm expands to 6.76 L by two
different pathways. Path A is an isothermal,
reversible expansion. Calculate the work for
Path A.
Answer in units of J
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To calculate the work for a reversible isothermal expansion, we can use the equation:
Work = -nRT ln(Vfinal/Vinitial)
Where:
- Work is the work done by the gas
- n is the number of moles of gas
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the absolute temperature in Kelvin
- Vinitial is the initial volume of the gas
- Vfinal is the final volume of the gas
First, let's calculate the number of moles of gas. We can use the ideal gas law equation:
PV = nRT
Rearranging the equation, we get:
n = PV / RT
Given:
P = 3.3 atm
Vinitial = 3.78 L
T = 318 K
Plugging in the values:
n = (3.3 atm * 3.78 L) / (0.0821 atm·L/(mol·K) * 318 K)
n = 0.4048 mol
Now, we can calculate the work using the formula:
Work = - (0.4048 mol * 0.0821 atm·L/(mol·K) * 318 K) ln(6.76 L / 3.78 L)
Calculating this expression:
Work = - (0.4048 mol * 0.0821 atm·L/(mol·K) * 318 K) ln(1.786 L)
Work = - (10.545 J/K) ln(1.786)
Using a calculator, the natural logarithm of 1.786 is approximately 0.5832:
Work = - (10.545 J/K) * 0.5832
Therefore, the work for Path A is approximately -6.14 J.