A cylinder with movable piston contains 0.1 mole od a monatomic ideal gas. The g

Question

A cylinder with movable piston contains 0.1 mole od a monatomic ideal gas. The gas, initially at state a, can be taken through either of two cycles, abca, or abcda,

Qc>a =685 J along the curved path

Wc>a=120 J along the curved path

Ua-Ub=450J

Wa>b>c=75J

a) Determine the change in internal energy, Ua-Uc, between states a and c

b)Is heat added or removed from the gas when the gas is taken along the path abc?

    caluclate the amount added or removed.

c)How much work is done on the gas in the process cda?

d)Is heat added or removed from the gas whe the gas is taken along the path cda?

   Explain your reasoning. +++Please show steps to answer but keep it simple and to the point:)

Explanation / Answer

a) According to 1st law of thermodynamics the change of internal energy moving from one state to another is equal to the work done on plus the heat transferred to the gas. Hence ?U_ca = U_a - U_c = W_ca + Q_ca = -120J + 685J = 565J b) Is there a typo? In you source the questioned path is abc For the path abc Internal energy is a state function. That means that the change of internal energy between two states is always the same irrespective of the path. So we know the difference between a and b: ?U_abc = U_c - U_a = -565J The work done on the path abc is given as: W_abc = 75J Apply 1st law of thermodynamics to find the heat transferred on the path: U_c - U_a = W_abc + Q_abc Q_abc = (U_c - U_a) - W_bca = -565J - 75J = -640J negative sign indicates that means the calculated amount is removed from the gas. For the path bca ?U_bca = U_a - U_b = 450J The work done on the path is: W = - ? p dV (from initial volume to final volume) = - p · ?V (for constant pressure process) - from b ? c W_bc = - p · (V_c - V_b ) = -3 × 10+5Pa · (0.75×10-3m³ - 1×10-3m³) = 75J - from c ? a W_ca = -120J (as given in the question => W_bca = W_bc + W_ca = -45J Hence: U_a - U_b = W_bca + Q_bca Q_bca = (U_a - U_b) - W_bca = 450J - (-45J) = 495J On this path heat is added to the gas c) The change of internal energy along the path is: ?U_cda = U_a - U_c = 565J So the internal energy rises. Since volume decreases along the path, we've got some expansion work, that means the work done on the path takes some negative value. Because the internal energy increases, Q need to be positive along the path, that means heat is added to the gas. The work done on the path is: - c ? d W_cd = 0 because the volume does not change - d ? a W_da = - p · (V_a - V_b ) = - 6×10+5Pa · (1.0×10-3m³ - 0.75×10-3m³) = -150J => W_cda = W_cd + W_da = -150J Hence: U_a - U_c = W_cda + Q_cda Q_cda = (U_a - U_c) - W_cda = 565J - (-150J) = 715J

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