A perfect gas (0.1 mol) having Cv = 1.50 R (independent of temperature) undergoes the following transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is (-) J (nearest integer) [Given: R = 0.082 L atm K-1 mol-1] [figure]
Answer
304
Solution
Step 1 to 2: isochoric (V = 1000 cm3)
w12 = 0
Step 2 to 3: isobaric at p = 3.00 atm, V: 1000 2000 cm3
w23 = -p V = -3.00 (2.000 - 1.000) L atm
= -3.00 L atm
Step 3 to 4: isochoric (V = 2000 cm3)
w34 = 0
w = w12 + w23 + w34 = -3.00 L atm
1 L atm = 101.3 J
w = -3.00 101.3 = -303.9 J -304 J
(The value of Cv is not needed: work depends only on the path in the p–V diagram.)
Answer: 304