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Chapter 6: Problem 34

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm}\). In the process, there is a heat gain by the system of \(350 . \mathrm{J} .\) c. A piston expands against \(1.00\) atm of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.

Short Answer

Expert verified
The internal energy changes for each scenario are: a) ΔU = 77 J b) ΔU ≈ 3855 J c) ΔU ≈ 384 J

Step by step solution

01

Setting up the equation

First, we need to set up the first law of thermodynamics equation: \(ΔU = q + W\)
02

Input the given values

In this case, we know that 100 Joules of work is required to compress the gas (W = 100 J) and the gas releases 23 Joules of heat (q = -23 J). Plug the values into the equation: \(ΔU = (-23) + (100)\)
03

Solve for internal energy change

Now, we just need to solve for ΔU: \(ΔU = 77 \mathrm{~J}\) So the internal energy change for scenario a is 77 Joules. #b. Calculating internal energy change when a piston is compressed against constant pressure and there is heat gain by the system#
04

Calculating work done on the system

The work done on the system can be calculated by using the formula: \(W = -PΔV\) Where: - P = Pressure (in atm) - ΔV = Change in volume (in L) We will convert the pressure and volume to SI units (Pascals and m³). Using the given values, we have, \(P = 1.9 \mathrm{~atm} = 1.9*101325 Pa\) and volume change, from \(8.3 L\) to \(2.8 L\), which gives us a change in volume, \(ΔV = 2.8 - 8.3 = -5.5 \mathrm{~L}\), which can be converted to cubic meters: \(-5.5\times10^{-3}m^3\). Now, we can find W: \(W = -1.9*101325 *(-5.5\times10^{-3})\)
05

Determine the internal energy change

Using the first law of thermodynamics formula, and substituting the given heat gain (\(q = 350 J\)) and the calculated value for work, we can find the internal energy change: \(ΔU = q + W\) \(ΔU = 350 + (-1.9*101325 *(-5.5\times10^{-3}))\)
06

Solve for internal energy change

Now, we just need to solve for ΔU: \(ΔU ≈ 3855 \mathrm{~J}\) So the internal energy change for scenario b is approximately 3855 Joules. #c. Calculating internal energy change when a piston expands against a constant pressure and heat is absorbed#
07

Calculating work done by the system

Similar to step 1 in solution b, we will calculate the work using \(W = -PΔV\). The given values are: \(P = 1.00 \mathrm{~atm} = 1.00*101325 Pa\) and, volume change, from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\), which gives us a change in volume, \(ΔV = 29.1 - 11.2 = 17.9 \mathrm{~L}\), which can be converted to cubic meters: \(17.9\times10^{-3}m^3\). Now, we can find W: \(W = -1.00*101325*(17.9\times10^{-3})\)
08

Determine the internal energy change

Using the first law of thermodynamics formula, and substituting the given heat absorbed value (\(q = 1037 J\)) and the calculated value for work, we can find the internal energy change: \(ΔU = q + W\) \(ΔU = 1037 + (-1.00*101325*(17.9\times10^{-3}))\)
09

Solve for internal energy change

Now, we just need to solve for ΔU: \(ΔU ≈ 384 \mathrm{~J}\) So the internal energy change for scenario c is approximately 384 Joules.

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Most popular questions from this chapter

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is \(40 . \mathrm{cm}^{3}\). If the combustion of this mixture releases 950. J of energy, to what volume will the gases expand against a constant pressure of 650 . torr if all the energy of combustion is converted into work to push back the piston?

Consider the following equations: $$ \begin{array}{rlrl} 3 \mathrm{~A}+6 \mathrm{~B} \longrightarrow & 3 \mathrm{D} & & \Delta H=-403 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{E}+2 \mathrm{~F} & \Delta H & =-105.2 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C} & \mathrm{A} & & \Delta H= & 64.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Suppose the first equation is reversed and multiplied by \(\frac{1}{6}\), the second and third equations are divided by 2 , and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \text { (unbalanced) } $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{~kJ}\). If \(2.50 \times 10^{2} \mathrm{~mL} \mathrm{~N}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50 \mathrm{~atm}\) and \(4.50 \times\) \(10^{2} \mathrm{~mL} \mathrm{O}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50\) atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

Consider the following statements: "Heat is a form of energy, and energy is conserved. The heat lost by a system must be equal to the amount of heat gained by the surroundings. Therefore, heat is conserved." Indicate everything you think is correct in these statements. Indicate everything you think is incorrect. Correct the incorrect statements and explain.

Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?
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