A heat engine with 0.500mol of a monatomic ideal gas initially fills a 1000cm^3

Question

A heat engine with 0.500mol of a monatomic ideal gas initially fills a 1000cm^3 cylinder at 700K . The gas goes through the following closed cycle: - Isothermal expansion to 5000cm^3 . - Isochoric cooling to 200K . - Isothermal compression to 1000cm^3 . - Isochoric heating to 700K A)How much work does this engine do per cycle? B)What is its thermal efficiency?

Explanation / Answer

(1) Work done BY the gas is given by the integral W = ? p dV from initial to final volume In the isochoric steps of the process no work is done, cause volume does no change, i.e W2 = W3 = 0 For an ideal gas undergoing an isothermal change of state: p·V = n·R·T = constant Hence: p·V = p_initial · V_initial p = p_initial · V_initial/V So the work done in thermal process is W = p_initial·V_initial ? 1/V dV from initial to final volume = p_initial·V_initial· ln(V_final/V_initial) Using ideal gas law, you can substitute p and V: p_initial·V_initial = n·R·T_initial Hence: W = n·R·T_initial · ln(V_final/V_initial) For the expansion process. W1 = 0.5mol · 8.3145J/molK · 800K · ln(6000cm³/1000cm³) = 5960J Work done by the gas in compression process is W3 = 0.5mol · 8.3145J/molK · 400K · ln(1000cm³/1000cm³) = -2980J So the net work done by the engine per cycle is: W = W1 + W2 + W3 + W4 = 5960J + 0J -2960J + 0J = 2960J 2) To calculate efficiency you need the heat transferred to the engine. You find it from internal energy considerations Change of internal energy of an ideal gas is given by: ?U = n·Cv·?T For a monatomic gas Cv = (3/2)·R On the other hand change of internal energy equals heat transferred to the gas minus work done By it (or plus work done on it): ?U = Q - W For the isothermal steps ?U = Q - W = 0 => Q = W For the isochoric steps W = 0 => Q = ?U = (3/2)·n·R·?T Hence: Q1 = W1 = 5960J Q2 = (3/2)·0.5mol·8.3145J/mol · (400K - 800K) = -2494J Q3 = W3 = -2980J Q4 = (3/2)·0.5mol·8.3145J/mol · (800K - 400K) = +2494J So the total heat transferred to the engine is Q_in = Q1+ Q4 = 5960J + 2494J = 8454J The total heat rejected by the engine is Q_out = - Q2 - Q3 = 2494J + 2980J = 5474J It efficiency is the ratio of net work done to heat absorbed: ? = W / Q_in = 2980J / 8454J = 0.352 = 35.2%

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