21. If 2.00 mol of an ideal gas at 300 K and 3.00 atm expands from 6.0 L to 12.0

Question

21. If 2.00 mol of an ideal gas at 300 K and 3.00 atm expands from 6.0 L to 12.0 L and a final pressure of 1.50 atm in two steps: (1) the gas is cooled at constant volume until its pressure has fallen to 1.20 atm, and (2) it is heated and allowed to expand against a constant pressure of 1.50 atm until its volume reaches 12.0 L, which of the following is correct? a) w=-2.74 kJ for the overall process b) w--1.82 kJ for step (1) and w -091 kJ for step (2) c) w=-3.46 kJ for the overall process d) w = 0 for step (I) and w =-912 J for step (2) e) w = 0 for the overall process 2 0 zt

Explanation / Answer

Step 1

Gas is cooled at constant volume

At constant volume

W1 = - P (V2-V1) = 0

step 2

gas is allowed to heated and expand

At constant pressure, P = 1.50 atm

W2 = - P (V2-V1)

= - 1.50 x (12 - 6)

= - 9 L-atm x 101.33 J/L-atm

= - 911.97 J

= - 912 J

Option D is the correct answer

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