6A (4) A sample of a monatomic ideal gas is taken through the process abca shown

Question

6A (4) A sample of a monatomic ideal gas is taken through the process abca shown below. At point a. the temperature is T 200 K P (N/m2) 20 a V (ms) (a) (5 points) What are the temperatures at points b and c? (b) (10 points) How much work is done by the gas during each step of the cycle? ab (c) (10 points) What was the net heat transferred to the gas during each step of the cycle? of Hint: The ideal gas equation allows you to compute the value of (n.R), and the First Law Thermodynamics gives you Qa,

Explanation / Answer

By PV = nRT

Moles of gas = 20*1/8.314*200 = 0.012 moles

Temp at b = 10N/m2*4m3/8.314JK-1mol-1*0.012mol = 400.93K

Temp at c = 10N/m2*1m3/8.314JK-1mol-1*0.012mol = 100.23K

Work ab = area under ab = 0.5*3*10 + 3*10 = 45J (expansion)

Work cb = area under bc = 3*10 = -30J (compression)

Work ca = area under ca = 0J

U=3/2nRT

Ubc = nCpdT = 0.012*2.5*8.314*300.7 = -75J

So Qbc = -75-(-30) = -45J

Uca = nCvdT = 0.012*1.5*8.314*99.77 = 14.93J

So Qca = 14.93J as dW = 0J

Since for cyclic process dU = 0 So Uab = -(-75+14.93) = 60.07J

So Qab = 60.07-45 = 15.07J

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.