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Chapter 6: Problem 39

A balloon filled with \(39.1\) moles of helium has a volume of \(876 \mathrm{~L}\) at \(0.0^{\circ} \mathrm{C}\) and \(1.00\) atm pressure. The temperature of the balloon is increased to \(38.0^{\circ} \mathrm{C}\) as it expands to a volume of \(998 \mathrm{~L}\), the pressure remaining constant. Calculate \(q, w\), and \(\Delta E\) for the helium in the balloon. (The molar heat capacity for helium gas is \(20.8 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{mol}\).)

Short Answer

Expert verified
In summary, for the helium in the balloon, we found the heat (q) to be \(31091.2 \mathrm{~J}\), the work (w) to be \(-12361.9 \mathrm{~J}\), and the change in internal energy (ΔE) to be \(18729.3 \mathrm{~J}\).

Step by step solution

01

Calculate the heat (q)

To find the heat (q), we can use the molar heat capacity (Cp) of helium gas, which is given as 20.8 𝑱/°C·mol. Since q = nCpΔT, where n is the number of moles and ΔT is the change in temperature, we can start by calculating ΔT: ΔT = T_final - T_initial = (38.0 - 0.0) °C = 38.0 °C Now, using the given moles (39.1 moles) and molar heat capacity, we can calculate the heat (q): q = (39.1 mol) × (20.8 J/°C·mol) × (38.0°C) = 31091.2 J
02

Calculate the work (w)

To find the work (w), we can use the formula: w = -PΔV, where P is the pressure and ΔV is the change in volume. First, convert the pressure to proper units: 1 atm = 101.3 kPa = 101325 Pa Now we can calculate ΔV as follows: ΔV = V_final - V_initial = (998 - 876) L = 122 L = 0.122 m³ Then, we can find the work (w): w = -PΔV = -(101325 Pa) × (0.122 m³) = -12361.9 J
03

Calculate the change in internal energy (ΔE)

To find the change in internal energy (ΔE), we can use the First Law of Thermodynamics, which states that ΔE = q + w. We have already calculated both q and w above, so we can plug these values in the formula: ΔE = q + w = (31091.2 - 12361.9) J = 18729.3 J In summary, the heat (q) is 31091.2 J, the work (w) is -12361.9 J, and the change in internal energy (ΔE) is 18729.3 J.

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Most popular questions from this chapter

When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.

Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) & \longrightarrow 4 \mathrm{PCl}_{3}(g) & & \Delta H=-1225.6 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & & \Delta H=-2967.3 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) & & \Delta H=-84.2 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{3} \mathrm{PO}(g) & & \Delta H=-285.7 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$

The sun supplies energy at a rate of about \(1.0\) kilowatt per square meter of surface area \((1\) watt \(=1 \mathrm{~J} / \mathrm{s})\). The plants in an agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{~m}^{2}\right)\). Assum- ing that sucrose is produced by the reaction \(12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+12 \mathrm{O}_{2}(g)\) \(\Delta H=5640 \mathrm{~kJ}\) calculate the percentage of sunlight used to produce the sucrose - that is, determine the efficiency of photosynthesis.

Consider the substances in Table 6.1. Which substance requires the largest amount of energy to raise the temperature of \(25.0 \mathrm{~g}\) of the substance from \(15.0^{\circ} \mathrm{C}\) to \(37.0^{\circ} \mathrm{C} ?\) Calculate the energy. Which substance in Table \(6.1\) has the largest temperature change when \(550 . \mathrm{g}\) of the substance absorbs \(10.7 \mathrm{~kJ}\) of energy? Calculate the temperature change.

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