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Chapter 6: Problem 42

As a system increases in volume, it absorbs 52.5 \(\mathrm{J}\) of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.500 \(\mathrm{atm} .\) The final volume of the system is 58.0 \(\mathrm{L}\) . What was the initial volume of the system if the internal energy of the system decreased by 102.5 \(\mathrm{J} ?\)

Short Answer

Expert verified
The initial volume of the system was approximately \(61.059 \mathrm{L}\).

Step by step solution

01

Convert pressure to Pascals (SI unit)

We have the pressure given in atm, so we need to convert it to Pascals (Pa). The conversion is 1 atm = 101325 Pa. So, \(P = 0.500 \mathrm{atm} \times \frac{101325 \mathrm{Pa}}{1 \mathrm{atm}} = 50662.5 \mathrm{Pa}\)
02

Apply the first law of thermodynamics equation

Now, let's rewrite the ΔU equation, plugging in the values we have: \(-102.5 \mathrm{J} = 52.5 \mathrm{J} - W\) We need to find the work done (W) to solve for the change in volume (ΔV).
03

Calculate work done

To find the work done (W), we can rearrange the equation we got in Step 2: \(W = 52.5 \mathrm{J} - (-102.5 \mathrm{J}) = 155 \mathrm{J}\)
04

Calculate change in volume (ΔV)

We can now use the formula for W to find the change in volume: \(W = -PΔV\) So, \(ΔV = -\frac{W}{P} = -\frac{155 \mathrm{J}}{50662.5 \mathrm{Pa}}\) To convert the volume in cubic meters (m³) to liters (L), we use the conversion factor 1 m³ = 1000 L. \(ΔV = -\frac{155 \mathrm{J}}{50662.5 \mathrm{Pa}} \times \frac{1000 \mathrm{L}}{1 \mathrm{m^3}} = -3.059 \mathrm{L}\)
05

Calculate the initial volume

We are given the final volume (58.0 L) and we have found the change in volume (ΔV = -3.059 L). We can find the initial volume (V₀) using the following equation: \(V₀ = V_\text{final} - ΔV\) So, \(V₀ = 58.0 \mathrm{L} - (-3.059 \mathrm{L}) = 61.059 \mathrm{L}\) The initial volume of the system was approximately 61.059 L.

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Most popular questions from this chapter

A swimming pool, 10.0 \(\mathrm{m}\) by \(4.0 \mathrm{m},\) is filled with water to a depth of 3.0 \(\mathrm{m}\) at a temperature of \(20.2^{\circ} \mathrm{C}\) . How much energy is required to raise the temperature of the water to \(24.6^{\circ} \mathrm{C} ?\)

Calculate \(\Delta E\) for each of the following. a. \(q=-47 \mathrm{kJ}, w=+88 \mathrm{kJ}\) b. \(q=+82 \mathrm{kJ}, w=-47 \mathrm{kJ}\) c. \(q=+47 \mathrm{kJ}, w=0\) d. In which of these cases do the surroundings do work on the system?

Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for $\mathrm{C}_{4} \mathrm{H}_{4}(-2341 \mathrm{kJ} / \mathrm{mol}), \mathrm{C}_{4} \mathrm{H}_{8}\( \)(-2755 \mathrm{kJ} / \mathrm{mol}),\( and \)\mathrm{H}_{2}(-286 \mathrm{kJ} / \mathrm{mol}),\( calculate \)\Delta H$ for the reaction $$ \mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) $$

It has been determined that the body can generate 5500 \(\mathrm{kJ}\) of energy during one hour of strenuous exercise. Perspiration is the body's mechanism for eliminating this heat. What mass of water would have to be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise? (The heat of vaporization of water is 40.6 \(\mathrm{kJ} / \mathrm{mol.} )\)

For each of the following situations a-c, use choices i-iii to complete the statement: "The final temperature of the water should be.." i. between \(50^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\) . ii. \(50^{\circ} \mathrm{C}\) . iii. between \(10^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\) . a. 100.0 -g sample of water at \(90^{\circ} \mathrm{C}\) is added to a 100.0 -g sample of water at \(10^{\circ} \mathrm{C}\) . b. A 100.0 -g sample of water at \(90^{\circ} \mathrm{C}\) is added to a $500.0 . \mathrm{g}\( sample of water at \)10^{\circ} \mathrm{C} .$ c. You have a Styrofoam cup with 50.0 \(\mathrm{g}\) of water at $10^{\circ} \mathrm{C}\( . You add a 50.0 -g iron ball at \)90^{\circ} \mathrm{C}$ to the water.
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