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Chapter 18: Problem 26

The internal energy of a gas is \(500 .\) J. The gas is compressed adiabatically, and its volume decreases by \(100 . \mathrm{cm}^{3} .\) If the pressure applied on the gas during compression is \(3.00 \mathrm{~atm},\) what is the internal energy of the gas after the adiabatic compression?

Short Answer

Expert verified
Answer: The final internal energy of the gas after the adiabatic compression is 530.4 J.

Step by step solution

01

Understand the First Law of Thermodynamics

The First Law of Thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat (Q) added to the system minus the work (W) done by the system, or: ΔU = Q - W For an adiabatic process, there is no heat exchange between the system and its surroundings, which means Q = 0. Therefore, the First Law of Thermodynamics for an adiabatic process can be simplified to: ΔU = - W
02

Calculate the work done during compression

The work (W) done during compression can be found using the formula: W = P * ΔV where P is the pressure applied on the gas during compression and ΔV is the change in volume. The given pressure is in atmospheres, but we need to convert it to Pascals (SI unit). To do this, we use the conversion factor 1 atm = 101325 Pa: P = 3.00 atm * 101325 Pa / 1 atm = 303975 Pa ΔV is given as -100 cm³. We need to convert it to m³ (SI unit) using the conversion factor 1 cm³ = 1e-6 m³: ΔV = -100 cm³ * 1e-6 m³ / 1 cm³ = -1e-4 m³ Now, we can calculate the work done during compression: W = 303975 Pa * -1e-4 m³ = -30.3975 J
03

Calculate the change in internal energy

Now that we know the work done during compression, we can use the First Law of Thermodynamics for adiabatic processes to calculate the change in internal energy: ΔU = - W = 30.3975 J
04

Calculate the final internal energy

Finally, we can calculate the internal energy of the gas after the adiabatic compression. The initial internal energy is given as 500 J. Final internal energy (U_final) = Initial internal energy (U_initial) + Change in internal energy (ΔU) U_final = 500 J + 30.3975 J = 530.3975 J The internal energy of the gas after the adiabatic compression is 530.4 J (rounded to one decimal place).

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Most popular questions from this chapter

A cryogenic storage container holds liquid helium, which boils at \(4.2 \mathrm{~K}\). Suppose a student painted the outer shell of the container black, turning it into a pseudoblackbody, and that the shell has an effective area of \(0.50 \mathrm{~m}^{2}\) and is at \(3.0 \cdot 10^{2} \mathrm{~K}\). a) Determine the rate of heat loss due to radiation. b) What is the rate at which the volume of the liquid helium in the container decreases as a result of boiling off? The latent heat of vaporization of liquid helium is \(20.9 \mathrm{~kJ} / \mathrm{kg} .\) The density of liquid helium is \(0.125 \mathrm{~kg} / \mathrm{L}\).

In which type of process is no work done on a gas? a) isothermal b) isochoric c) isobaric d) none of the above

Which of the following does not radiate heat? a) ice cube b) liquid nitrogen c) liquid helium d) a device at \(T=0.010 \mathrm{~K}\) e) all of the above f) none of the above

Suppose \(0.010 \mathrm{~kg}\) of steam (at \(100.00^{\circ} \mathrm{C}\) ) is added to \(0.10 \mathrm{~kg}\) of water (initially at \(\left.19.0^{\circ} \mathrm{C}\right)\). The water is inside an aluminum cup of mass \(35 \mathrm{~g}\). The cup is inside a perfectly insulated calorimetry container that prevents heat flow with the outside environment. Find the final temperature of the water after equilibrium is reached.

For a class demonstration, your physics instructor pours \(1.00 \mathrm{~kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) over \(4.00 \mathrm{~kg}\) of ice at \(0.00^{\circ} \mathrm{C}\) and allows the system to reach equilibrium. He is then going to measure the temperature of the system. While the system reaches equilibrium, you are given the latent heats of ice and steam and the specific heat of water: \(L_{\text {ice }}=3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg}\), \(L_{\text {steam }}=2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}, c_{\text {water }}=4186 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right) .\) You are asked to calculate the final equilibrium temperature of the system. What value do you find?
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