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

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)

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

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)