Question

PVT Cycle Due never 7.5 Pressure (kN/m2) 2.5 a с 1.0 3.0 Volume (mº) A sample of an ideal gas is taken through the cyclic process shown on the p-v diagram in the figure. The temperature of the gas at point a is 120 K. How many moles of gas are in the sample? Submit Answer Tries 0/16 What is the temperature in kelvins at point b? Submit Answer Tries 0/16 What is the temperature in kelvins at point c? Submit Answer Tries 0/16 What is the net heat added to the gas during the cycle in joules? Submit Answer Tries 0/16 Post Discussion Send Feedback

Answer 1

Hello Dear,

Given, P (KM) 3 A N/m² a c 52.5 X 10 va = 1 m3 ļ 3 7.5810 N/m² v (m) es vo 3 m3 Pe 2:5X103 N/m2 = 3 m² V # is n, Ta = 120k No. of moles present in the gas PV = nRTA R= 8.314 J/molak 2 na Pava 215x18x1 RT 8:314 X120 We have aa 응 n = 2.5 #. Temperature at point b, To= P, V, 7.5 x163x3 NR To=1082.5K 2.5X8.314 #. is om isochoric process (v = long tant) isobaric (p= constant) So, for the process eta(p=constant) bc

VOT Te 38/to July > Te ve Ta = 3x120 => Te = 360k Va # We so know from the law of thermodynamics od Q = dut ofw But for the cyclic process, the system returns to its initial state. As the internal energy is a "state functions du=o da=dw Now, the net heat added to the cycle = net work done. Area under the cycle { x (ve - V) x (Po – Pe) Ź x 2 x (7.5-2.5) X103 Net heat added 5000 J 3
Hope this helps..