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Chapter 7: Problem 61

There are other forms of work besides \(\mathrm{P}-\mathrm{V}\) work. For example, electrical work is defined as the potential \(x\) change in charge, \(w=\phi d q\). If a charge in a system is changed from \(10 \mathrm{C}\) to \(5 \mathrm{C}\) in a potential of \(100 \mathrm{V}\) and \(45 \mathrm{J}\) of heat is liberated, what is the change in the internal energy? (Note: \(1 \mathrm{V}=1 \mathrm{J} / \mathrm{C})\).

Short Answer

Expert verified
The change in internal energy is -455J.

Step by step solution

01

Calculate the work done by the system

The work is calculated using \(W=\phi dq\). Here, \(\phi\) is 100V and \(dq\) is the change in charge, which is \(10C - 5C = 5C\). So, the work \(W = 100V * 5C = 500J\). According to the convention, work done by the system is positive.
02

Calculate the change in internal energy

Applying the first law of thermodynamics, \(\Delta U=Q-W\). Here, \(Q = 45J\) (the heat liberated) and \(W = 500J\) (the work done by the system). So, \(\Delta U = 45J - 500J = -455J\). The change in internal energy is thus -455J.

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Most popular questions from this chapter

A 1.22 kg piece of iron at \(126.5^{\circ} \mathrm{C}\) is dropped into \(981 \mathrm{g}\) water at \(22.1^{\circ} \mathrm{C} .\) The temperature rises to \(34.4^{\circ} \mathrm{C} .\) What will be the final temperature if this same piece of iron at \(99.8^{\circ} \mathrm{C}\) is dropped into \(325 \mathrm{mL}\) of glycerol, \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}(1)\) at \(26.2^{\circ} \mathrm{C} ?\) For glycerol, \(d=1.26 \mathrm{g} / \mathrm{mL} ; C_{n}=219 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

A plausible final temperature when \(75.0 \mathrm{mL}\) of water at \(80.0^{\circ} \mathrm{C}\) is added to \(100.0 \mathrm{mL}\) of water at \(20^{\circ} \mathrm{C}\) is (a) \(28^{\circ} \mathrm{C} ;\) (b) \(40^{\circ} \mathrm{C} ;\) (c) \(46^{\circ} \mathrm{C} ;\) (d) \(50^{\circ} \mathrm{C}\)

A \(1.00 \mathrm{g}\) sample of \(\mathrm{Ne}(\mathrm{g})\) at 1 atm pressure and \(27^{\circ} \mathrm{C}\) is allowed to expand into an evacuated vessel of \(2.50 \mathrm{L}\) volume. Does the gas do work? Explain.

What will be the final temperature of the water in an insulated container as the result of passing \(5.00 \mathrm{g}\) of steam, \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\) at \(100.0^{\circ} \mathrm{C}\) into \(100.0 \mathrm{g}\) of water at \(25.0^{\circ} \mathrm{C} ?\left(\Delta H_{\mathrm{vap}}^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\right)\).

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) torches are used in welding. How much heat (in kJ) evolves when 5.0 L of \(C_{2} \mathrm{H}_{2}\) \(\left(d=1.0967 \mathrm{kg} / \mathrm{m}^{3}\right)\) is mixed with a stoichiometric amount of oxygen gas? The combustion reaction is $$\begin{array}{r} \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-1299.5 \mathrm{kJ} \end{array}$$
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