Question

p(N/m2) 20 10 1.0 x 10 V(m) 50 x 103 0 25x10.3 0 newtons per 1. One mole of an ideal monatomic gas, initially at point A at a pressure of 1.0 x 1 meter squared and a volume of 25 x 10 m', is taken through a 3-process cycle, as shown in the capacities for constant volume and constant pressure are, respectively, C /2)R (5/2)R, where R is the universal gas constant, 8.32 J/mole K. Determine each of the following: sibly. For a monatomic gas, the heat the temperature of the gas at each of the vertices, A, B. and C, of the triangular cycle b. the net work done by the gas for one cycle c. the net heat absorbed by the gas for one full cycle d. the heat given off by the gas for the third process from C to A e. the efficiency of the cycle

Answer 1

(a) Temperature of the gas at each of the vertices A, B and C of triangular cycle which will be given as -

using an ideal gas law ; P V = n R T

At vertices A, we have

T_A = P_A V_A / n R

T_A = [(1 x 10⁵ Pa) (25 x 10^-3 m³)] / [(1 mole) (8.32 J/mol.K)]

T_A = 300 K

At vertices B, we have

T_B = P_B V_B / n R

T_B = [(2 x 10⁵ Pa) (25 x 10^-3 m³)] / [(1 mole) (8.32 J/mol.K)]

T_B = 600 K

At vertices C, we have

T_C = P_C V_C / n R

T_C = [(1 x 10⁵ Pa) (55 x 10^-3 m³)] / [(1 mole) (8.32 J/mol.K)]

T_C = 600 K

(b) Net work done by the gas for one cycle which will be given as -

For cycle ABC, we have

A = (1/2) (base) (height)

A = (0.5) [(50 x 10^-3 m³) - (25 x 10^-3 m³)] [(2 x 10⁵ Pa) - (1 x 10⁵ Pa)]

A = [(0.5) (25 x 10^-3 m³) (1 x 10⁵ Pa)]

A = 1250 J

We know that, W_net = P $\Delta$V = (Area of cycle ABC)

W_net = 1250 J

(c) Net heat absorbed by the gas for one full cycle which will be given as -

According to first law of thermodynamic, we have

$\Delta$U = Q_net + W_net

Q_net = [(0 J) + (1250 J)]

Q_net = 1250 J

(d) Heat given off by the gas for third process from C to A which be given as -

For process C $\rightarrow$ A, we have

Q = m C_p$\Delta$T

Q = (1 kg) [(5/2) (8.32 J/mol.K)] (300 K)

Q = 6240 J

(e) The efficiency of cycle which will be given by -

_$\eta$ = W / Q_in

_$\eta$ = (1250 J) / [(1250 J) + (6240 J)]

_$\eta$ = [(1250 J) / (7490 J)]

_$\eta$ = 0.16