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Chapter 6: Problem 30

A system undergoes a process consisting of the following two steps: Step 1: The system absorbs \(72 \mathrm{~J}\) of heat while \(35 \mathrm{~J}\) of work is done on it. Step 2: The system absorbs \(35 \mathrm{~J}\) of heat while performing \(72 \mathrm{~J}\) of work. Calculate \(\Delta E\) for the overall process.

Short Answer

Expert verified
The change in internal energy (∆E) for the overall process is 144 J.

Step by step solution

01

Calculate ∆E for the first step.

Using the first law of thermodynamics: ∆E₁ = Q₁ - W₁ Here, Q₁ = 72 J (heat absorbed) and W₁ = 35 J (work done on the system). ∆E₁ = 72 J - 35 J = 37 J The change in internal energy for the first step is 37 J.
02

Calculate ∆E for the second step.

Using the first law of thermodynamics again: ∆E₂ = Q₂ - W₂ Here, Q₂= 35 J (heat absorbed) and W₂ = -72 J (since the system is performing work). ∆E₂ = 35 J - (-72 J) = 35 J + 72 J = 107 J The change in internal energy for the second step is 107 J.
03

Calculate the total ∆E for the overall process.

To find the total change in internal energy for the overall process, we can sum up the calculated ∆E values for each step: ∆E_total = ∆E₁ + ∆E₂ = 37 J + 107 J = 144 J The change in internal energy for the overall process is 144 J.

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Most popular questions from this chapter

Hess's law is really just another statement of the first law of thermodynamics. Explain.

A system releases \(125 \mathrm{~kJ}\) of heat while \(104 \mathrm{~kJ}\) of work is done on it. Calculate \(\Delta E\).

Consider the substances in Table 6.1. Which substance requires the largest amount of energy to raise the temperature of \(25.0 \mathrm{~g}\) of the substance from \(15.0^{\circ} \mathrm{C}\) to \(37.0^{\circ} \mathrm{C}\) ? Calculate the energy. Which substance in Table \(6.1\) has the largest temperature change when \(550 . \mathrm{g}\) of the substance absorbs \(10.7 \mathrm{~kJ}\) of energy? Calculate the temperature change.

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm} .\) In the process, there is a heat gain by the system of \(350 . \mathrm{J}\). c. A piston expands against \(1.00 \mathrm{~atm}\) of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.

Calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ \begin{aligned} 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) & \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1010 . \mathrm{kJ} \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-317 \mathrm{~kJ} \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-143 \mathrm{~kJ} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \Delta H &=-286 \mathrm{~kJ} \end{aligned} $$
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