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Chapter 15: Problem 59

An ideal gas is heated at a constant pressure of $2.0 \times 10^{5} \mathrm{Pa}\( from a temperature of \)-73^{\circ} \mathrm{C}$ to a temperature of \(+27^{\circ} \mathrm{C}\). The initial volume of the gas is $0.10 \mathrm{m}^{3} .\( The heat energy supplied to the gas in this process is \)25 \mathrm{kJ} .$ What is the increase in internal energy of the gas?

Short Answer

Expert verified
Answer: The increase in internal energy of the gas is 15 kJ.

Step by step solution

01

Convert temperatures to Kelvin

To do any calculations with temperatures, it's essential to convert them to Kelvin first. To convert a Celsius temperature to Kelvin, we simply add 273.15. Initial temperature in Kelvin: \(T_1 = -73^{\circ}\mathrm{C} + 273.15 = 200.15\ \mathrm{K}\) Final temperature in Kelvin: \(T_2 = 27^{\circ}\mathrm{C} + 273.15 = 300.15\ \mathrm{K}\)
02

Calculate the final volume of the gas

Since we are given the constant pressure, we can use the ideal gas law to find the final volume of the gas. The ideal gas law can be written as \(PV = nRT\) Since the pressure, number of moles n, and the gas constant R are constant, we can write \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) We are given the initial volume (\(V_1 = 0.10\ \mathrm{m}^3\)) and have found the initial and final temperatures in Kelvin. Now, we can solve for the final volume \(V_2\): \(V_2 = \frac{T_2}{T_1} \times V_1 = \frac{300.15\ \mathrm{K}}{200.15\ \mathrm{K}} \times 0.10\ \mathrm{m}^3 \approx 0.150\ \mathrm{m}^3\)
03

Calculate the work done by the gas

The work done by the gas can be calculated using the constant pressure and the change in volume: \(W = P \Delta V = P(V_2 - V_1) = (2.0 \times 10^5\ \mathrm{Pa}) (0.150\ \mathrm{m}^3 - 0.10\ \mathrm{m}^3) = 10\ \mathrm{kJ}\)
04

Use the First Law of Thermodynamics to find the increase in internal energy

The First Law of Thermodynamics states that \(\Delta U = Q - W\) Where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added, and \(W\) is the work done by the system. We are given the amount of heat energy supplied to the gas \(Q = 25\ \mathrm{kJ}\), and we have calculated the work done by the gas \(W = 10\ \mathrm{kJ}\). Now we can find the increase in internal energy: \(\Delta U = 25\ \mathrm{kJ} - 10\ \mathrm{kJ} = 15\ \mathrm{kJ}\) So, the increase in internal energy of the gas is 15 kJ.

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Most popular questions from this chapter

A coal-fired electrical generating station can use a higher \(T_{\mathrm{H}}\) than a nuclear plant; for safety reasons the core of a nuclear reactor is not allowed to get as hot as coal. Suppose that $T_{\mathrm{H}}=727^{\circ} \mathrm{C}\( for a coal station but \)T_{\mathrm{H}}=527^{\circ} \mathrm{C}$ for a nuclear station. Both power plants exhaust waste heat into a lake at \(T_{\mathrm{C}}=27^{\circ} \mathrm{C} .\) How much waste heat does each plant exhaust into the lake to produce \(1.00 \mathrm{MJ}\) of electricity? Assume both operate as reversible engines.
The internal energy of a system increases by 400 J while \(500 \mathrm{J}\) of work are performed on it. What was the heat flow into or out of the system?
A \(0.500-\mathrm{kg}\) block of iron at \(60.0^{\circ} \mathrm{C}\) is placed in contact with a \(0.500-\mathrm{kg}\) block of iron at $20.0^{\circ} \mathrm{C} .\( (a) The blocks soon come to a common temperature of \)40.0^{\circ} \mathrm{C} .$ Estimate the entropy change of the universe when this occurs. [Hint: Assume that all the heat flow occurs at an average temperature for each block.] (b) Estimate the entropy change of the universe if, instead, the temperature of the hotter block increased to \(80.0^{\circ} \mathrm{C}\) while the temperature of the colder block decreased to \(0.0^{\circ} \mathrm{C}\) [Hint: The answer is negative, indicating that the process is impossible. \(]\)
An engine releases \(0.450 \mathrm{kJ}\) of heat for every \(0.100 \mathrm{kJ}\) of work it does. What is the efficiency of the engine?
A town is planning on using the water flowing through a river at a rate of \(5.0 \times 10^{6} \mathrm{kg} / \mathrm{s}\) to carry away the heat from a new power plant. Environmental studies indicate that the temperature of the river should only increase by \(0.50^{\circ} \mathrm{C} .\) The maximum design efficiency for this plant is \(30.0 \% .\) What is the maximum possible power this plant can produce?
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