a heat engine takes 2.5 mol of an ideal gas through the reversible cycle abca, o

Question

a heat engine takes 2.5 mol of an ideal gas through the reversible cycle abca, on the PV diagram shown in the figure the path bc is an isothermal process. the temperature at C is 710k and the volumes at a and c are .018m^3 and .17m^3, respectively. the molar heat capacity at constant volume of the gas is CV=37 J /mol K and the ideal gas constant is R= 8.314 j/mol k find the following values

a cytle abca, a process. The temperature at c is 710 K, and the volumes at a and c are figure. The path bc is an isothermal process. The temperatur 0.018,m and 0.17,m', respectively. The molar and the ideal gas constant is R 8314 Wmol-K. find the followi heat capacity at constant volume, of the gas, is Cy -37 Wmokk. 17. PA a. 110900 Pa b. 47490 Pa d 86020,Pa e. 147700 Pab 153500 Pa 86810 Pa 18. PR a 426300 Pa b 1338000 Pa d. 785100 Pa e. 1117000 Pa f 819900 Pa Ve a. 119.3.K b. 38.33,K d. 44.3, K e 131.7,x f. 129.3.x 20, (must still find works and heats)

Explanation / Answer

First, I calculated the temperature and pressures at points a, b, and c using PV = nRT.

Point A:

V = 0.018 m^3

T = 710 K

P = (2.5)(8.31)(710) / 0.018 = 819458.33 Pa

Point C:

V = 0.17 m^3

T = 710 K

P = nRT / v = (2.5)(8.31)(710) / 0.17 = 86766.176 Pa

Point B:

V = 0.018 m^3

T = PV / NR = (86766.176)(0.018) / (2.5)(8.31) = 75.176 K

P = 86766.176 Pa

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