Final answer:
The change in internal energy (ΔE) of the gas during expansion is -271 J, and the change in enthalpy (ΔH) is -34 J calculated using the first law of thermodynamics and the given values for heat released and work done during expansion.
Step-by-step explanation:
To calculate the change in internal energy (ΔE) of a gas during expansion or compression, the first law of thermodynamics is used, which is expressed as ΔE = q + w, where q is the heat absorbed by the system, and w is the work done by the system. If heat is released by the gas, q is negative; if work is done by the gas (expansion), w is negative. In the problem provided, the gas releases 34 J of heat (q = -34 J), and it expands against a constant pressure. We can calculate work using the formula w = -PΔV, where P is the pressure in atmospheres and ΔV is the change in volume in liters.
The change in volume ΔV = final volume - initial volume = 5.2 L - 3.4 L = 1.8 L. Next, we convert the pressure from atm to J by using the conversion factor 1 atm = 101.325 J/L. Therefore, the work done by the gas during expansion is w = -1.30 atm × 1.8 L × 101.325 J/(L·atm) = -236.787 J.
Now we can find the change in internal energy using ΔE = q + w, which is ΔE = -34 J + (-236.787 J) = -270.787 J (which we can round to -271 J). Finally, the change in enthalpy (ΔH) for a process at constant pressure is given by ΔH = q, which means the change in enthalpy is ΔH = -34 J.