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Chapter 6: Problem 35

A sample of an ideal gas at \(15.0 \mathrm{~atm}\) and \(10.0 \mathrm{~L}\) is allowed to expand against a constant external pressure of \(2.00 \mathrm{~atm}\) at a constant temperature. Calculate the work in units of \(\mathrm{kJ}\) for the gas expansion. (Hint: Boyle's law applies.)

Short Answer

Expert verified
The work done by the gas during expansion is approximately \(-13.17 \mathrm{~kJ}\).

Step by step solution

01

1. Applying Boyle's Law

: Boyle's Law states that for a given amount of gas at constant temperature, the pressure of the gas multiplied by its volume is constant. Mathematically, this can be expressed as: \[P_1V_1 = P_2V_2\] where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume respectively. We are given the initial pressure and volume, and the final pressure (which is equal to the external pressure). We can use Boyle's Law to find the final volume.
02

2. Finding the final volume

: From Boyle's Law, \(P_1V_1 = P_2V_2\). Given, \(P_1 = 15.0 \mathrm{~atm}\), \(V_1 = 10.0 \mathrm{~L}\), and \(P_2 = 2.00 \mathrm{~atm}\). So, we can find the final volume as: \[V_2 = \frac{P_1V_1}{P_2} = \frac{(15.0 \mathrm{~atm})(10.0 \mathrm{~L})}{2.00 \mathrm{~atm}} = 75.0 \mathrm{~L}\]
03

3. Calculating work done by the gas

: We can calculate the work done by the gas during expansion as: \[W = -P_{ext}(V_2 - V_1)\] where \(W\) is the work done, \(P_{ext}\) is the external pressure, and \(V_1\) and \(V_2\) are the initial and final volumes. Given \(P_{ext} = 2.00 \mathrm{~atm}\), \(V_1 = 10.0 \mathrm{~L}\), and \(V_2 = 75.0 \mathrm{~L}\), we can plug in the values to get: \[W = -(2.00 \mathrm{~atm})(75.0 \mathrm{~L} - 10.0 \mathrm{~L})\]
04

4. Converting atm to J

: Before calculating the work done, we need to convert the pressure from atm to J. The conversion factor is 1 atm = 101.325 J/L. Thus: \[P_{ext} = 2.00 \mathrm{~atm} \times \frac{101.325 \mathrm{~J}}{1 \mathrm{~L}} = 202.65 \mathrm{~atm.L^{-1}}\]
05

5. Calculating the work done

: Now that we have the pressure in J/L, we can plug in the values into the work formula: \[W = -(202.65 \mathrm{~J/L})(65.0 \mathrm{~L}) = -13172.25 \mathrm{~J}\]
06

6. Converting the work done to kJ

: To convert the work done in J to kJ, we simply divide by 1000: \[W = -13.17 \mathrm{~kJ}\] The work done by the gas during expansion is approximately -13.17 kJ.

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Most popular questions from this chapter

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?

Which of the following substances have an enthalpy of formation equal to zero? a. \(\mathrm{Cl}_{2}(g)\) b. \(\mathrm{H}_{2}(g)\) c. \(\mathrm{N}_{2}(l)\) d. \(\mathrm{Cl}(g)\)

The reaction $$ \mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ is the last step in the commercial production of sulfuric acid. The enthalpy change for this reaction is \(-227 \mathrm{~kJ}\). In designing a sulfuric acid plant, is it necessary to provide for heating or cooling of the reaction mixture? Explain.

At \(298 \mathrm{~K}\), the standard enthalpies of formation for \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are \(227 \mathrm{~kJ} / \mathrm{mol}\) and \(49 \mathrm{~kJ} / \mathrm{mol}\), respectively. a. Calculate \(\Delta H^{\circ}\) for $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) $$ b. Both acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) can be used as fuels. Which compound would liberate more energy per gram when combusted in air?

Assume that \(4.19 \times 10^{6} \mathrm{~kJ}\) of energy is needed to heat a home. If this energy is derived from the combustion of methane \(\left(\mathrm{CH}_{4}\right)\), what volume of methane, measured at STP, must be burned? \(\left(\Delta H_{\text {combustion }}^{\circ}\right.\) for \(\mathrm{CH}_{4}=-891 \mathrm{~kJ} / \mathrm{mol}\) )
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