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Chapter 7: Problem 49

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.

Short Answer

Expert verified
Yes, the gas does work during the process. You can deduce this by identifying the initial and final states of the system, determining the change in volume, and then calculating the work done by the system. The work done as calculated will be negative, indicating that the gas did work on the surroundings during its expansion.

Step by step solution

01

Identify the initial state

Let's identify the initial state. Here, the \(1.00 \mathrm{g}\) sample of \(\mathrm{Ne}(\mathrm{g})\) at 1 atm pressure and\(27^{\circ} \mathrm{C}\) forms our initial state.
02

Identify the final state

Now let's identify the final state. When the gas is allowed to expand into an evacuated vessel of \(2.50 L\), this forms our final state.
03

Determine the change in volume

Since the initial state is in a container that is filled to a certain volume, and the final state is in an evacuated vessel of a known larger volume, the volume change \(\Delta V\) will be the final volume of the vessel.
04

Calculate the work done

Work done \(W\) by the gas can be calculated using the formula \( W = -P\Delta V\), where \(P\) is the pressure the gas initially had and \(\Delta V\) is the change in volume. Recall that the work done by gas on its surroundings in an isothermal (constant temperature) expansion is negative, as it uses its internal energy to push against the external pressure.
05

Conclusion

If the calculated work done is more than zero, the gas did work while expanding into the evacuated volume.

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Most popular questions from this chapter

How much heat, in kilojoules, is evolved in the complete combustion of (a) \(1.325 \mathrm{g} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ;\) (b) \(28.4 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(\mathrm{STP} ;(\mathrm{c})\) \(12.6 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(23.6^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg} ?\) Assume that the enthalpy change for the reaction does not change significantly with temperature or pressure. The complete combustion of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) is represented by the equation $$\begin{array}{r} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-2877 \mathrm{kJ} \end{array}$$

A sample gives off 5228 cal when burned in a bomb calorimeter. The temperature of the calorimeter assembly increases by \(4.39^{\circ} \mathrm{C} .\) Calculate the heat capacity of the calorimeter, in kilojoules per degree Celsius.

Determine \(\Delta H^{\circ}\) for this reaction from the data below. \(\mathrm{N}_{2} \mathrm{H}_{4}(1)+2 \mathrm{H}_{2} \mathrm{O}_{2}(1) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(1)\) $$\begin{array}{r} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-622.2 \mathrm{kJ} \end{array}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(1) \quad \Delta H^{\circ}=-187.8 \mathrm{kJ}$$

Some of the butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) in a \(200.0 \mathrm{L}\) cylinder at \(26.0^{\circ} \mathrm{C}\) is withdrawn and burned at a constant pressure in an excess of air. As a result, the pressure of the gas in the cylinder falls from 2.35 atm to 1.10 atm. The liberated heat is used to raise the temperature of 132.5 L of water in a heater from 26.0 to 62.2 ^ C. Assume that the combustion products are \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) exclusively, and determine the efficiency of the water heater. (That is, what percent of the heat of combustion was absorbed by the water?)

The heat of solution of \(\mathrm{KI}(\mathrm{s})\) in water is \(+20.3 \mathrm{kJ} / \mathrm{mol}\) KI. If a quantity of KI is added to sufficient water at \(23.5^{\circ} \mathrm{C}\) in a Styrofoam cup to produce \(150.0 \mathrm{mL}\) of 2.50 M KI, what will be the final temperature? (Assume a density of \(1.30 \mathrm{g} / \mathrm{mL}\) and a specific heat of \(2.7 \mathrm{Jg}^{-1}\) \(\left.^{\circ} \mathrm{C}^{-1} \text {for } 2.50 \mathrm{M} \mathrm{KI} .\right)\)
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