Question

Calculate the work done when 1.000 mole of an ideal gas expands reversibly from 1.0 L to 10 L at 298.0 K. Then, calculate the amount of work done when 1.000 mole of an ideal gas at 298.0 K expands irreversibly against a constant external pressure of 1.00 atm. Compare the two values and comment. The question continues at the bottom.

Wrev=-nRtln(vf/vi)

-(1.000 mol)(0.08205 L*atm/mol*K)(298K)ln(10L/1L).

=-(56.3 L*atm) (101.325 J/ L*atm)

=-5705J

My question is where do the values 56.3 L*atm and 101.325 J/ L*atm come from, what are they derived from? When I multiplied -(1.000 mol)(0.08205 L*atm/mol*K)(298K)ln(10L/1L) I get -129.63 L*atm/K.This was a completed example I was given in class and I couldn't break it down.

Expert Answer

Answer 1

You forget to compute the calculation systematically.

101.325 J/L*atm is obtained by dividing R in J/(mol*K) with R in L*atm/(mol*K)

Step-by-step explanation:

Fo the given equation

Wrev = -nRtln(vf/vi)

We have the calculation below:

Wrev = -(1.000 mol)(0.08205 L*atm/mol*K)(298K)ln(10L/1L)

Wrev = -24.4509*ln(10) = -24.4509*2.303 = -56.3 L*atm

However, the unit was converted from L*atm to J by using a conversion unit derived from the ratio of gas constant (R) in J/(mol*K) and L*atm/(mol*K).

R = 0.082057 L*atm/(mol*K)

R = 8.3145 J/(mol*K)

Therefore, we can derive a conversion unit for L*atm to J as:

[8.3145 J/(mol*K)]/[0.082057 L*atm/(mol*K)] = 101.325 J/L*atm

Thus:

Wrev = -56.3 (L*atm) * 101.325 (J/L*atm) = -5705 J