Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 1
ID: 1903506 • Letter: C
Question
Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 1 and 2 in the figure below. In process 1 the gas is heated at constant volume from an initial pressure of Pi = 95 kPa to a final pressure of Pf = 190 kPa. In process 2 the gas expands at constant pressure from an initial volume of Vi = 0.85 m3 to a final volume of Vf = 2.55 m3. (a) How much heat is added to the gas during these two processes? process 1 - 120 kJ process 2 - 810 kJ (b) How much work does the gas do during this expansion? (c) What is the change in the internal energy of the gas?Explanation / Answer
Q: Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 3 and 4 in the figure below. In process 3 the gas expands at constant pressure from an initial volume of Vi = 0.80 m3 to a final volume of Vf = 2.40 m3. In process 4 the gas is heated at constant volume from an initial pressure of Pi = 103 kPa to a final pressure of Pf = 206 kPa. http://www.webassign.net/walker/18-27alt.gif (a) How much heat is added to the gas during these two processes? process 3 _______ kJ process 4 _____ jK (b) How much work does the gas do during this expansion? Which i was able to find and it was 165 kJ (c) What is the change in the internal energy of the gas?' ANS: (a) Process 3 is a constant pressure process. For such a process the heat added to a system equals its change in enthalpy. Q = ?H The enthalpy of an ideal gas is given by: H = n·Cp·T So the change in enthalpy for a constant amount is: ?H = n·Cp·?T For a monatomic ideal gas, which has molar heat capacity at constant volume of Cp = (5/2)·R it is Q = ?H = (5/2)·n·R·?T Using ideal gas you can express Q in terms of pressure and volume n·R·T = p·V => n·R·?T = ?(p·V) = p·?V (because p is constant) Hence, Q = ?H = (5/2)·p·?V = (5/2) · p_i·(V_f - V_i) = (5/2) · 103 kPa · (2.40 m³ - 0.80 m³) = 412 kJ Process 4 is a constant volume process. For such a process the heat added to a system equals its change in in internal energy. Q = ?U With U = n·Cv·T and Cv = (3/2)·R for a monatomic ideal gas follows Q = ?U = (3/2)·n·R·?T From ideal gas law follows that n·R·?T = ?(p·V) = V·?p because V is constant in this process Hence, Q = ?U = (3/2)·V·?p = (3/2) · V_f·(p_f - p_i) = (3/2) · 2.4 m³ · (206 kPa - 103 kPa) = 370.8 kJ (b) Work done to surrounding is given by: W = ? p dV form initial to final volume In process 4 the volume is constant so no work is done For constant pressure process 3 the work integral simplifies to: W = p · ? dV = p·?V = p_i · (V_f - V_i) = 103 kPa · (2.40 m³ - 0.80 m³) = 164.8 kJ (c) Total change in internal energy equals total heat added to the gas minus the work it does to the surrounding: ?U = ?Q - ?W = 412 kJ + 370.8 kJ - 164.8 kJ = 618 kJ