Question

A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When 0.470 kJ of heat is added to the gas, it expands and does 219 J of work on the surroundings.

1. What is the value of ΔH for this process?Express the energy in kilojoules to three significant digits.

2. What is the value of ΔE for this process? Express the energy in kilojoules to three decimal places.

Answer 1

According to first law of thermodynamics, U= q- w or U=q-pv

q= heat absorbed by the system

w is the work done =219 J, convert to kJ by diviing with 1000.

So first find U = 0.470 kJ- (219x10^-3 kJ) =0.470-0.219 = 0.251 kJ

Now, enthalpy change, H= U+ pV

here, pV is the work done (w)

=0.251 kJ+0.219 = 0.470 kJ

2.E is the energy change

So total enthalpy change is the 0.470 kJ and before did a work of 219 J or 0.219 kJ, so to find the change in energy it will be

E = 0.47-0.219= 0.251kJ