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Chapter 6: Problem 18

The work done to compress a gas is \(74 \mathrm{~J}\). As a result, \(26 \mathrm{~J}\) of heat is given off to the surroundings. Calculate the change in energy of the gas.

Short Answer

Expert verified
The change in energy of the gas is \(48 \mathrm{~J}\).

Step by step solution

01

Identifying the given quantities

Let's identify the quantities we are given. We are told that the work done to compress the gas is \(74 \mathrm{~J}\) (J for Joules, the unit for work and energy). Furthermore, the gas gives off \(26 \mathrm{~J}\) of heat energy to its surroundings.
02

Understanding the energy conservation principle

The energy conservation applies, which means any work done on the gas goes into increasing its internal energy or is lost as heat energy. The formula to use for the change in energy of the gas is the work done minus the heat given off. The formula is \(\Delta E = W - Q\), where \( \Delta E \) is the change in energy, \(W\) is the work done, and \(Q\) is the heat energy given off. The energy units in this problem are Joules.
03

Calculating the change in energy

To find the change in energy of the gas, we substitute the given values into the equation: \(\Delta E = 74 \mathrm{~J} - 26 \mathrm{~J}\). This simplifies to \(\Delta E = 48 \mathrm{~J}\). So the change in energy of the gas is \(48 \mathrm{~J}\).

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Most popular questions from this chapter

Calculate the standard enthalpy change for the reaction $$ 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ given that $$ \begin{aligned} 2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow & \mathrm{Al}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow & \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-822.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

The internal energy of an ideal gas depends only on its temperature. Do a first-law analysis of this process. A sample of an ideal gas is allowed to expand at constant temperature against atmospheric pressure. (a) Does the gas do work on its surroundings? (b) Is there heat exchange between the system and the surroundings? If so, in which direction? (c) What is \(\Delta E\) for the gas for this process?

On what law is the first law of thermodynamics based? Explain the sign conventions in the equation \(\Delta E=q+w\)

A 2.10 -mole sample of crystalline acetic acid, initially at \(17.0^{\circ} \mathrm{C}\), is allowed to melt at \(17.0^{\circ} \mathrm{C}\) and is then heated to \(118.1^{\circ} \mathrm{C}\) (its normal boiling point) at 1.00 atm. The sample is allowed to vaporize at \(118.1^{\circ} \mathrm{C}\) and is then rapidly quenched to \(17.0^{\circ} \mathrm{C}\), so that it recrystallizes. Calculate \(\Delta H^{\circ}\) for the total process as described.

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is an organic solvent and is also used as a fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol: $$ \begin{aligned} 2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-1452.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$
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