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

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.

Answer 2

Answer:

$\Delta E=21.34kJ$

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:

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

So we compute the energy change:

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

Regards.