The P-V m of a cyclic thermal p for 2 g of He is shown below. The system follows

Question

The P-V m of a cyclic thermal p for 2 g of He is shown below. The system follows the path abcda. The volume of the gas at or c is 0.025 m The pressures at points a, b, c, d can be easily from the graph as well as the volumes at a read and d. The process ab is an isothermal process The temperature at the apex a is 1202.79 K. The molecular mass of He is 4 g/mole. Treat ideal monoatomic gas. a) Find the number of moles of He gas. b) Find the temperature of the gas at apex d in Kelvins. c) Find the internal energy change during the constant volume process da.

Explanation / Answer

a)

Ideal gas equation PV = nRT

At point a:

P = 5.0*10^5 Pa

V = 0.010 m^3

R = 8.314 J/ mole K

T = 1202.79 K

number of moles = 5.0*10^5*0.010/(8.314*1202.79) = 0.5 moles.

b)

from a to d, the volume is constant hence the temperature is directly proportional to pressure.

Temperature = T_1*P_2/P_1 = 1202.79*1/5 = 240.558 K

c) Internal energy change = nCvdeltaT = n*(3R/2)*delta T = 1.5*n R delta T = 1.5*0.5*8.314*(240.558-1202.79) = -5999.997 J which is approximately -6000 J.

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