Figure shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. The temperature of gas at c and d are 300 and 500 K. The heat absorbed by the gas during the process is given by K (100) R λ n 2. Find the value of K.
Text Solution
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(4)
Change in internal energy for cyclic process
( Δ U) = 0
for process a → b, (P = constant)
W a → b = P Δ V = nR Δ T = – 400 R
for process b → c, (T = constant)
W b → c = – 2R (300) λ n2
for process c → d (P = constant)
W c → d = + 400R
for process d → a, (T = constant)
W d → a = 2R (500) λ n2
Δ W = W a → b + W b → c + W c → d + W d → a
Δ W = 400R λ n2
Δ Q = Δ W
Δ Q = 400 R λ n 2 = 4(100) R λ n2
∴ K = 4
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