Respuesta :
Explanation:
- For path A, the calculation will be as follows.
As, for reversible isothermal expansion the formula is as follows.
W = [tex]-2.303 nRT log(\frac{V_2}{V_1})[/tex]
Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.
As the given data is as follows.
R = 8.314 J/(K mol), T = 298 K ,
[tex]V_{2}[/tex] = 8.34 L, [tex]V_{1}[/tex] = 2.67 L
Now, putting the given values into the above formula as follows.
W = [tex]-2.303 nRT log(\frac{V_2}{V_1})[/tex]
= [tex]-2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})[/tex]
= [tex]-2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494[/tex]
= -2818.68 J
Hence, work for path A is -2818.68 J.
- For path B, the calculation will be as follows.
Step 1: When there is no change in volume then W = 0
Hence, for step 1, W = 0
Step 2: As, the gas is allowed to expand against constant external pressure [tex]P_{external}[/tex] = 1.00 atm.
So, W = [tex]-P_{external} \times \Delta V[/tex]
Now, putting the given values into the above formula as follows.
W = [tex]-P_{external} \times \Delta V[/tex]
= [tex]-1 atm \times (8.34 L - 2.67 L)[/tex]
= -5.67 atm L
As we known that, 1 atm L = 101.33 J
Hence, work will be calculated as follows.
W = [tex]-\frac{101.33 J}{1 atm L} \times 5.67 atm L[/tex]
= -574.54 J
Therefore, total work done by path B = 0 + (-574.54 J)
W = -574.54 J
Hence, work for path B is -574.54 J.
