2. A gas absorbs 21.39 kJ of energy while expanding against 0.276 atm from a volume of 0.0432 L to 1.876 L. What is the energy change of the gas?

In this case, given the first law of thermodynamics in which the absorbed heat by the system and the work done are related including the energy change as shown below:

[tex]Q-W=\Delta E[/tex]

We can compute the work by knowing the final and initial volumes:

[tex]W=P\Delta V=0.276atm\times (1.876L-0.0432L)\\\\W=0.506atm*L*\frac{101.325kPa}{1atm}*\frac{1m^3}{1000L}=0.0513kJ[/tex]

So we compute the energy change:

[tex]\Delta E=21.39kJ-0.0513kJ\\\\\Delta E=21.34kJ[/tex]

Regards.

oeerivona oeerivona · Answer 2 · 2020-05-28T02:48:34+02:00

Answer:

The energy change of the gas is 21338.74 J

Explanation:

Here we have;

Work done, w by the gas = p×dV

Where:

p = pressure of the gas = 0.276 atm = 27965.7 Pa

dV = Change in volume = V₂ - V₁

V₂ = Final volume of the gas = 1.876 L

V₁ = Initial volume of the gas = 0.0432 L

Therefore, dV = 1.876 - 0.0432 = 1.8328 L = 0.0018328 m³

w = 27965.7×0.0018328 = 51.26 J

Therefore, the energy change = Energy absorbed - Work done

Energy absorbed = 21.39 kJ = 21390 J

Hence, the energy change of the gas = 21390 - 51.26 = 21338.74 J.