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\(\bullet\) An experimenter adds 970 \(\mathrm{J}\) of heat to 1.75 \(\mathrm{mol}\) of an ideal gas to heat it from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C}\) at constant pressure. The gas does \(+223 \mathrm{Jof}\) work during the expansion. (a) Calculate the change in internal energy of the gas. (b) Calculate \(\gamma\) for the gas.

Short Answer

Expert verified
(a) ΔU = 747 J; (b) γ = 1.30.

Step by step solution

01

Identify Known Values

For the given problem, we are told that the heat added ( Q) is 970 J, the work done by the gas (W) is 223 J, and the temperature change is from 10.0°C to 25.0°C.
02

Apply First Law of Thermodynamics

The First Law of Thermodynamics states ΔU = Q - W, where ΔU is the change in internal energy. We substitute the given values: Q = 970 J and W = 223 J. Thus, ΔU = 970 J - 223 J = 747 J.
03

Determine Specific Heat Capacity at Constant Pressure

The change in temperature, ΔT, is 25.0°C - 10.0°C = 15.0°C or 15.0 K (since the change in temperature is the same in Celsius and Kelvin). The formula for heat added at constant pressure is Q = nC_pΔT, where n is the number of moles, C_p is the molar heat capacity, and ΔT is the temperature change. We know Q = 970 J and n = 1.75 mol, so C_p = Q / (nΔT) = 970 J / (1.75 mol × 15.0 K) = 37.14 J/mol·K.
04

Determine Specific Heat Capacity at Constant Volume

Using the relationship C_v = ΔU / (nΔT), we can determine the molar heat capacity at constant volume. We have ΔU = 747 J, n = 1.75 mol, and ΔT = 15.0 K, thus C_v = 747 J / (1.75 mol × 15.0 K) = 28.46 J/mol·K.
05

Calculate Gamma ( γ) for the Gas

The adiabatic index ( γ) is the ratio of the heat capacities, γ = C_p / C_v. Substituting in our values, we find γ = 37.14 J/mol·K / 28.46 J/mol·K = 1.30.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Law of Thermodynamics
The First Law of Thermodynamics is a pivotal principle in understanding energy changes within a system. It states that the total energy in an isolated system is constant, implying energy cannot be created or destroyed but can change forms. This principle can be mathematically represented as: \[ \Delta U = Q - W \] where
  • \( \Delta U \) is the change in internal energy
  • \( Q \) is the heat added to the system
  • \( W \) is the work done by the system
In our problem, the experimenter added 970 J of heat (\( Q \)) to the system, and the gas performed 223 J of work (\( W \)). Thus, the change in internal energy \( \Delta U \) was calculated as 747 J. This is a straightforward application of the First Law, illustrating how energy transfers and transformations balance within the gas.
Ideal Gas
An ideal gas is a theoretical gas that perfectly follows the ideal gas laws without any deviations. It's an important concept because it provides a simplified but very useful model of how gases behave under different conditions. The characteristics of an ideal gas include:
  • Particles that have negligible size
  • No intermolecular forces between particles, meaning interactions only occur during collisions
  • Elastic collisions, meaning kinetic energy is conserved
Ideal gases are defined by the equation of state known as the Ideal Gas Law:\[ PV = nRT \]where
  • \( P \) is the pressure
  • \( V \) is the volume
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
In the given problem, the ideal gas assumption allows us to directly relate the heat added and work done to the change in internal energy, using simplified relations valid for ideal gases.
Internal Energy
Internal energy is the total energy contained within a system, originating from the microscopic movements and interactions of its molecules. For an ideal gas, this internal energy primarily depends on temperature because:- It encompasses kinetic energy of particles, which is directly related to temperature.- The formula for change in internal energy of an ideal gas at constant volume can be simplified to: \[ \Delta U = nC_v \Delta T \]In this scenario, the gas's internal energy increased by 747 J when heated from 10.0°C to 25.0°C. This change was determined using the First Law of Thermodynamics, confirming that the internal energy change was due to both the heat added and the work performed by the gas during expansion.
Specific Heat Capacity
Specific heat capacity, a critical concept in thermodynamics, refers to the amount of heat required to change the temperature of a substance per unit mass. When dealing with gases, we often use molar heat capacities:
  • \( C_p \): specific heat capacity at constant pressure
  • \( C_v \): specific heat capacity at constant volume
For our ideal gas, these were calculated as follows:- \( C_p = \frac{Q}{n\Delta T} \approx 37.14 \text{ J/mol·K} \)- \( C_v = \frac{\Delta U}{n\Delta T} \approx 28.46 \text{ J/mol·K} \)The ratio of these two, known as the adiabatic index (\( \gamma \)), was found to be 1.30. This index is significant because it provides insights into how gases expand and compress without heat exchange, influencing the speed of sound in the gas and other dynamic properties.

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Most popular questions from this chapter

\(\bullet$$\bullet\) STP. The conditions of standard temperature and pressure (STP) are a temperature of \(0.00^{\circ} \mathrm{C},\) and a pressure of 1.00 atm. (a) How many liters does 1.00 mol of any ideal gas occupy at STP? (b) For a scientist on Venus, an absolute pressure of 1 Venusian-atmosphere is 92 Earth-atmospheres. Of course she would use the Venusian atmosphere to define STP. Assuming she kept the same temperature, how many liters would 1 mole of ideal gas occupy on Venus?

\(\bullet$$\bullet$$\bullet\) A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. (The calculation of the buoyant force is discussed in Chapter \(13 .\) ) If the volume of the balloon is 500.0 \(\mathrm{m}^{3}\) and the surrounding air is at \(15.0^{\circ} \mathrm{C},\) what must the temperature of the air in the balloon be for it to lift a total load of 290 \(\mathrm{kg}\) (in addition to the mass of the hot air)? The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.23 \(\mathrm{kg} / \mathrm{m}^{3}.\) BIO Temperature and degrees of freedom. The internal energy of an ideal monatomic gas is simply the kinetic energy associated with the translational motion of its atoms as they move randomly in each of the three independent spatial dimensions. However, for a diatomic ideal gas we must also take into account the kinetic and potential energies associated with molecular vibration, and the kinetic energy associated with molecular rotation. Roughly speaking, each independent way that energy can be stored is known as a degree of freedom. Although translational motion can occur at any temperature, rotational and vibrational motions typically cannot occur at lower temperatures- thus, the number of available degrees of freedom can change as the temperature changes. For example, an ideal monatomic gas has three degrees of freedom (one for each of its independent directions of translational motion) at all temperatures. In contrast, diatomic hydrogen \(\left(\mathrm{H}_{2}\right)\) has five degrees of freedom near room temperature \((3\) translational and 2 rotational). However, at higher temperatures, where molecular vibrations can occur, diatomic hydrogen has seven degrees of freedom \((3\) translational, 2 rotational, and 2 vibrational). The equipartition theorem states that equilibrium each degree of freedom contributes \(\frac{1}{2} n R T\) to the internal energy of the gas. For example, the internal energy of a monatomic gas, which has 3 degrees of freedom, would be \(\frac{3}{2} n R T\).

\(\bullet$$\bullet\) A cylinder 1.00 \(\mathrm{m}\) tall with inside diameter 0.120 \(\mathrm{m}\) is used to hold propane gas (molar mass 44.1 \(\mathrm{g} / \mathrm{mol} )\) for use in a barbecue. It is initially filled with gas until the gauge pressure is \(1.30 \times 10^{6} \mathrm{Pa}\) and the temperature is \(22.0^{\circ} \mathrm{C} .\) The temperature of the gas remains constant as it is partially emptied out of the tank, untill the gauge pressure is \(2.50 \times 10^{5}\) Pa. Calculate the mass of propane that has been used.

\(\bullet\) A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains 499 \(\mathrm{cm}^{3}\) of air at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right)\) and a temperature of \(27.0^{\circ} \mathrm{C}\) At the end of the stroke, the air has been compressed to a volume of 46.2 \(\mathrm{cm}^{3}\) and the gauge pressure has increased to \(2.72 \times 10^{6} \mathrm{Pa} .\) Compute the final temperature.

\(\bullet\) A 20.0 \(\mathrm{L}\) tank contains 0.225 \(\mathrm{kg}\) of helium at \(18.0^{\circ} \mathrm{C}\) . The molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol} .\) (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

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