Problem 7 115 points] The path less taken. (treated as an ideal monatomic gas) i
ID: 1778097 • Letter: P
Question
Problem 7 115 points] The path less taken. (treated as an ideal monatomic gas) is held at T 300 K and P-10 atm. The One mole of argon (treated as an ideal monatomic gas) is held at Ti-300 K and P, -10 atm. The gas is then allowed to expand to P 1 atm along a variety of pathways. For each pathway specified below, calculate () the final temperature Th (G) the work done w, (ii) the heat flow q (iv) the change in energy and (v) the change in enthalpy /. (a) A reversible, isothermal expansion. (b) An adiabatic expansion against a constant external pressure of 1 atm. (c) An adiabatic expansion into a vacuum.Explanation / Answer
REVERSIBLE ISOTHERMAL EXPANSION
final temperature = initial temperature = 300 K
Work done = -2.303nRT log(Pi/Pf) = -2.303 x 1 x 0.082057 (L atm mol-1 K-1) x300x log(10/1) = 57.0043 L atm
Heat flow = work done, because change in internal energy is zero in an isothermal reversible process
Change in internal energy =0 An ideal gas by definition has no interactions between particles, no intermolecular forces, so pressure change at constant temperature does not change internal energy.
Enthalpy H = U +PV
H = U + nRT = 0
Change in enthalpy is zero for an isothermal reversible expansion