ThermodynamicsHard
Question
One mole of a non-ideal gas undergoes a change of state (2 atm, 3 L, 95 K) → (4 atm, 5 L, 245 K) with a change in internal energy, ΔU = 30.0 L-atm. The change in enthalpy (ΔH) of the process in L-atm is
Options
A.40.0
B.42.3
C.44.0
D.undefined, because pressure is not constant
Solution
$\Delta H = \Delta U = \Delta(PV) = \Delta U + \left( P_{2}V_{2} - P_{1}V_{1} \right)$
$= 30.0\text{ L-atm + }(4 \times 5 - 2 \times 3)\text{ L-atm} $$$\text{=44.0 L-atm}$$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
1 mol of NH3 gas at 27oC is expanded under adiabatic condition to make volume 8 times (g = 1.33). Final temperature and ...An ideal gas in a thermally insulated vessel at internal pressure = P1, volume = V1 and absolute temperature = T1 expand...The normal boiling point of water is 100°C. At 100°C...For which of the following change ᐃH ≠ ᐃE ?...The change that does not increase entropy is...