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Chapter 7: Problem 39

A balloon filled with \(39.1 \mathrm{moles}\) of helium has a volume of \(876 \mathrm{L}\) at \(0.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}\) 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
The heat (q) absorbed by the helium gas is \(30872.56 \mathrm{J}\), the work (w) done by the gas is \(-12361.7 \mathrm{J}\), and the change in internal energy (ΔE) of the gas is \(18510.86 \mathrm{J}\).

Step by step solution

01

Convert temperatures to Kelvin

Since we will be working with the Ideal Gas Law and temperature changes, it's important to convert the given temperatures from Celsius to Kelvin. Initial temperature, \(T_1 = 0.0^{\circ} \mathrm{C} = 273.15 \mathrm{K}\) Final temperature, \(T_2 = 38.0^{\circ} \mathrm{C} = 273.15 + 38.0 = 311.15 \mathrm{K}\)
02

Calculate the change in volume

To find the work done by the helium gas, we need to find the change in volume. ΔV = final volume - initial volume ΔV = \(998 \mathrm{L} - 876 \mathrm{L} = 122 \mathrm{L}\)
03

Calculate the work done

We know the following equation for work done by a gas when the pressure is constant: w = -PΔV In this case, the constant pressure P is given as \(1.00 \mathrm{atm}\). Since we are given heat capacity in J/°C, we should change the pressure unit from atm to Pascal. \(1 \mathrm{atm} = 101325 \mathrm{Pa}\) So, w = \(-1.00 \times 101325 \mathrm{Pa} \times 122 \mathrm{L}\) Convert litres to cubic meters: \(1\mathrm{L} = 0.001\mathrm{m^3}\) w = \(-101325 \mathrm{Pa} \times 0.122\mathrm{m^3} = -12361.7 \mathrm{J}\) The work done by the helium gas is -12361.7 J.
04

Calculate the heat absorbed by the gas

We can use the following equation for the heat absorbed for a constant pressure process: q = nCΔT Here, n = 39.1 moles, C = 20.8 J/mol°C, and ΔT = T2 - T1 = 311.15 K - 273.15 K = 38.0 K. So, q = \(39.1 \mathrm{mol} \times 20.8 \frac{\mathrm{J}}{\mathrm{mol\cdot K}} \times 38.0\mathrm{K} = 30872.56\mathrm{J}\) The heat absorbed by the helium gas is 30872.56 J.
05

Calculate the change in internal energy

Now that we have both q and w, we can use the first law of thermodynamics to find ΔE: ΔE = q + w ΔE = \(30872.56 \mathrm{J} - 12361.7 \mathrm{J} = 18510.86 \mathrm{J}\) The change in internal energy of the helium gas is 18510.86 J. Solution: The heat (q) absorbed by the helium gas is 30872.56 J, the work (w) done by the gas is -12361.7 J, and the change in internal energy (ΔE) of the gas is 18510.86 J.

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Most popular questions from this chapter

In a coffee-cup calorimeter, \(150.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl is added to \(50.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) to make \(200.0 \mathrm{g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\). If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{kJ} / \mathrm{mol},\) calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is \(4.184 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) and assume no heat loss to the surroundings.

The enthalpy change for the reaction $$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ is \(-891 \mathrm{kJ}\) for the reaction \(a s\) written. a. What quantity of heat is released for each mole of water formed? b. What quantity of heat is released for each mole of oxygen reacted?

If the internal energy of a thermodynamic system is increased by \(300 .\) \(\mathrm{J}\) while \(75 \mathrm{J}\) of expansion work is done, how much heat was transferred and in which direction, to or from the system?

What is the difference between \(\Delta H\) and \(\Delta E ?\)

For the reaction \(\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\frac{1}{2} \mathrm{O}_{2}(g), \Delta H=+90.7 \mathrm{kJ}\). a. What quantity of heat is required to produce 1 mole of mercury by this reaction? b. What quantity of heat is required to produce 1 mole of oxygen gas by this reaction? c. What quantity of heat would be released in the following reaction as written? $$2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{HgO}(s)$$
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