Question

2. First Law of Thermodynamics An engine takes 3.25 mole of an ideal monatomic He gas through the cycle shown in the figure. Note that the temperature of the gas does not change during process c-a. p(Pa X 105) a b 2.0 - V(m3) 0 0.010 0.040 Hints: Cp = Cv+ R. Monatomic and diatomic gases have a different number of degrees of freedom at intermediate temperatures like those in this problem. 1[Pa] = 1 [N/m2] is the metric unit of pressure. a) Find the pressure at point a. b) Find the temperature at points a, b, and c. c) How much heat went in or out of the gas during segments a-b, b-c, and C-a? In each case explain whether heat entered or left the gas.

Answer 1

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Given there is 3.25 moles of an ideal mono atomic gas P Given c-a is isothermal. P 2.0*10 Pa 0.040 m 0.010 m. 3. 2.0 Nc + 0.040 0.010 ✓ → Va = frem ist law_of_thermodynamics. we can write that whese Due change AU + W in_internal energy We amount of work done Qe quantity of heart (Pa) poessúre at a) We need to find the point We knows_PV = nRT Since cto a is an isothermal process, temperatures at - both_the_points will be the same. So, we can write ? Pavaa Pave Pave & Now substitute the values. Va.. . Pa (0.040). 8x10² Pa (0.01) Pa : S 5 x 10 Temperatuse at paints, b, &c [To To and te ] We can solve this the ideal using gas equaturnale ele PlanRT

Palao nR Ta Ta :, Pava 8x10?x (0.01) (35),(3:314) NR. Ta= 296.07K Po Vb .. Similarly n R Pb Ub = nRTO To= 8x105 (0.640] 3:05 *8-314) 1184.29% of Note 3 isobaric Pb=Pa=8x105 Pa a to b is an as process Now Tope Pe la 2x 108 x 10.040 296:07K HR H 3.25'x 8.3.14 at We álkeady know that Tata Tas.cto arisan. Isothermal pisacess We need to calculate the heat went in or out.. i a-b =) process, pressure luore semains It's an isobaric Constant: AU+Who Q We know WL PAI Avi and Au nCAT @ NCAT_+pAV .

We know that the maboatomic gas has specific heat Capacity lat constant volume as ve Eye 3 R 3 R. . Qe na RAT + pov [184.28=296.01 AU. 2.25 x 3 x 8.314 * SU 36000.22 J pave 8x10*x[0.040- 0.010] 84,900J AU + W 36000:22 + 24,000J Q. + 60000.22 J Cheat comes in Q istre . ü b-c it's an isochovic procesie e volume demains.constant. So, out W Q- n SVAT+ pow Since volume is constant, AVEO re QE DU nCAT 2) 3.25x3 Rx: 0296.02 - W=O1 1184.29 geta [catio 305*328314x [216.02- 184-24 = - 36000. 22J ;-) heat went out Q is event and Custo]

E C-a - It's an iso thermal process , as Q, Dot W NGAT + W. AT=o temperature remains constant nou DT = 0; ie DU=0 So, Win an isothermal process ocess is w z ART In / VA vi Q: W az nRT ln In la 0.010 Q = 3.25 x 8-314x tn x 296.07 0.040 Q 2 -11090 29 J heat went out