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Chapter 6: Problem 36

A system absorbs 35 \(\mathrm{J}\) of heat and has 25 \(\mathrm{J}\) of work performed on it. The system then returns to its initial state by a second step. If 5 \(\mathrm{J}\) of heat are given off in the second step, how much work is done by the system in the second step?

Short Answer

Expert verified
The work done by the system in the second step is \(5 \, J\).

Step by step solution

01

Calculate the change in internal energy for step 1

For the first step, we are given the heat absorbed (Q_step1) and work performed on the system (W_step1). We can use these values to find the change in internal energy during the first step: \(∆U_{step1} = Q_{step1} - W_{step1}\) where \(Q_{step1} = 35 \, J\) (heat absorbed during step 1) \(W_{step1} = 25 \, J\) (work performed on the system during step 1) Plug the given values into the equation: \(∆U_{step1} = 35 \, J - 25 \, J = 10 \, J\)
02

Calculate the change in internal energy for step 2

In the second step, the system returns to its initial state, so the total change in the internal energy for both steps is zero: \(\Delta U_{total} = \Delta U_{step1} + \Delta U_{step2} = 0\) Since we have calculated the change in the internal energy for step 1, we can rearrange this equation to find the change in the internal energy for step 2: \(∆U_{step2} = -∆U_{step1}\) Plug in the change in the internal energy for step 1: \(∆U_{step2} = -(10 \, J) = -10 \, J\)
03

Calculate the work done by the system in the second step

We are given the heat given off in the second step (Q_step2) and can use the change in the internal energy for step 2 to find the work done by the system in the second step (W_step2): \( ∆U_{step2} = Q_{step2} - W_{step2} \) Rearrange the equation to solve for W_step2: \(W_{step2} = Q_{step2} - ∆U_{step2} \) where \(Q_{step2} = -5 \, J\) (heat given off during step 2) Plug in the given values into the equation: \(W_{step2} = -5 \, J - (-10 \, J) = 5 \, J\) So, the work done by the system in the second step is 5 J.

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Most popular questions from this chapter

A gas absorbs 45 kJ of heat and does 29 kJ of work. Calculate \(\Delta E .\)

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)(\text { unbalanced }) $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{kJ}\) . If $2.50 \times 10^{2} \mathrm{mL} \mathrm{N}_{2}(g)\( at \)100 .^{\circ} \mathrm{C}$ and 3.50 atm and \(4.50 \times\) \(10^{2} \mathrm{mL} \mathrm{O}_{2}(g)\) at $100 .^{\circ} \mathrm{C}$ and 3.50 atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

A coffee-cup calorimeter initially contains 125 \(\mathrm{g}\) water at \(24.2^{\circ} \mathrm{C} .\) Potassium bromide \((10.5 \mathrm{g}),\) also at \(24.2^{\circ} \mathrm{C},\) is added to the water, and after the KBr dissolves, the final temperature is \(21.1^{\circ} \mathrm{C}\) . Calculate the enthalpy change for dissolving the salt in \(\mathrm{J} / \mathrm{g}\) and $\mathrm{kJ} / \mathrm{mol}$ . Assume that the specific heat capacity of the solution is 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) and that no heat is transferred to the surroundings or to the calorimeter.

The combustion of 0.1584 g benzoic acid increases the temperature of a bomb calorimeter by \(2.54^{\circ} \mathrm{C}\) . Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is 26.42 \(\mathrm{kJ} / \mathrm{g} .\) ) A \(0.2130-\mathrm{g}\) sample of vanillin \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) is then burned in the same calorimeter, and the temperature increases by $3.25^{\circ} \mathrm{C}$ . What is the energy of combustion per gram of vanillin? Per mole of vanillin?

What is the difference between \(\Delta H\) and \(\Delta E ?\)
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