Question

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?

Answer 1

The energy change of the gas is 21338.74 J

Step-by-step 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.

Answer 2

`\Delta E=21.34kJ`

Step-by-step explanation:

Hello,

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:

`Q-W=\Delta E`

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

`W=P\Delta V=0.276atm* (1.876L-0.0432L)\\\\W=0.506atm*L*(101.325kPa)/(1atm)*(1m^3)/(1000L)=0.0513kJ`

So we compute the energy change:

`\Delta E=21.39kJ-0.0513kJ\\\\\Delta E=21.34kJ`

Regards.