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The temperature of \(2.00 \mathrm{~mol}\) of an ideal monatomic gas is raised \(15.0 \mathrm{~K}\) in an adiabatic process. What are (a) the work \(W\) done by the gas, (b) the energy transferred as heat \(Q,(\mathrm{c})\) the change \(\Delta E_{\text {int }}\) in internal energy of the gas, and (d) the change \(\Delta K\) in the average kinetic energy per atom?

Short Answer

Expert verified
(a) Work done \(W = -373.41 \mathrm{~J}\); (b) \(Q = 0\); (c) \(\Delta E_{\text{int}} = 373.41 \mathrm{~J}\); (d) \(\Delta K = 3.11 \times 10^{-22} \mathrm{~J}\).

Step by step solution

01

Understand the Adiabatic Process

In an adiabatic process, no heat is transferred in or out of the system. Therefore, the energy transferred as heat \(Q\) is zero. This means that the change in the internal energy \(\Delta E_{\text{int}}\) of the system equals the work \(W\) done on or by the system.
02

Change in Internal Energy

The change in internal energy \(\Delta E_{\text{int}}\) of an ideal gas can be determined using the formula \(\Delta E_{\text{int}} = \frac{3}{2} n R \Delta T\), where \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(\Delta T\) is the change in temperature. Here, \(n = 2.00 \), \(\Delta T = 15.0 \mathrm{~K}\), and \(R = 8.314 \mathrm{~J/(mol \cdot K)}\). Substituting these values, we get: \[ \Delta E_{\text{int}} = \frac{3}{2} \times 2.00 \times 8.314 \times 15.0 \].
03

Calculate the Work Done by the Gas

Since \(Q = 0\) (adiabatic process), the work done \(W\) by the gas is equal and opposite to the change in internal energy because \(W = -\Delta E_{\text{int}}\). Once we compute \(\Delta E_{\text{int}}\) from Step 2, the work done \(W = -\Delta E_{\text{int}}\).
04

Determine Change in Kinetic Energy per Atom

For a monatomic ideal gas, the average kinetic energy per atom is given by \(K = \frac{3}{2}kT\), where \(k\) is the Boltzmann constant. The change in kinetic energy per atom \(\Delta K\) can be expressed as \(\frac{3}{2}k\Delta T\). Calculate this using \(k = 1.38 \times 10^{-23} \mathrm{~J/K}\) and \(\Delta T = 15.0 \mathrm{~K}\).
05

Final Calculations

Compute the results from the formulas: \(\Delta E_{\text{int}} = \frac{3}{2} \times 2.00 \times 8.314 \times 15.0 = 373.41 \mathrm{~J}\). Thus, \(W = -373.41 \mathrm{~J}\). The change in kinetic energy per atom is \(\Delta K = \frac{3}{2} \times 1.38 \times 10^{-23} \times 15.0 = 3.11 \times 10^{-22} \mathrm{~J}\).

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Key Concepts

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

Ideal Monatomic Gas
An ideal monatomic gas refers to a gas composed of individual atoms rather than molecules. The term 'ideal' implies that the gas follows the ideal gas laws, where the particles move independently and do not exert forces on each other except during elastic collisions.
The characteristics of an ideal monatomic gas include:
  • Individual atoms have negligible volume compared to the container.
  • There are no attractive or repulsive forces between the atoms.
  • The gas obeys the equation of state: PV = nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature.
Ideal monatomic gases, like helium or neon, are perfect for exploring kinetic theory and thermodynamics as their simple nature offers clear insight into energy transformations.
Internal Energy Change
The internal energy of a gas is the total energy of all its atoms or molecules due to their motion (kinetic energy) and positions (potential energy). In an ideal monatomic gas, the internal energy is entirely kinetic since there are no interactions between atoms.
For a change in internal energy (\( \Delta E_{\text{int}} \) ), we use the expression:\[\Delta E_{\text{int}} = \frac{3}{2} n R \Delta T\]where \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( \Delta T \) is the change in temperature.
This equation shows how internal energy changes with temperature in an ideal gas. During an adiabatic process, all energy change arises from temperature change, as there's no heat exchange with the environment.
Work Done by Gas
Work done by a gas in a thermodynamic process relates to the energy transferred as the gas expands or contracts. In an adiabatic process, this work is manifested through changes in internal energy because the system is insulated from heat exchange. Thus, the work done \( W \) by the gas is described by:\[W = -\Delta E_{\text{int}}\]since any energy decrease in the system corresponds to work done on surroundings.
In our example, since the internal energy change is 373.41 J, the work done by the gas is -373.41 J, indicating that the energy was used for doing work on the environment without any heat gain or loss.
Kinetic Energy Change
The average kinetic energy per atom in an ideal monatomic gas is crucial for understanding temperature effects. This energy determines how fast the atoms move. For a single atom's average kinetic energy \( K \), the formula is:\[K = \frac{3}{2}kT\]where \( k \) is the Boltzmann constant and \( T \) is the temperature.
During a temperature change, this becomes \( \Delta K = \frac{3}{2} k \Delta T \), which allows us to calculate changes in kinetic energy due to a temperature shift.
In a rise of 15.0 K, for example, each atom's kinetic energy changes by \( \Delta K = 3.11 \times 10^{-22} \text{ J} \). Understanding these changes sheds light on energy dynamics and temperature relations in an ideal gas environment.

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Most popular questions from this chapter

We know that for an adiabatic process \(p V^{\gamma}=\) a constant. Evaluate "a constant" for an adiabatic process involving exactly \(2.0 \mathrm{~mol}\) of an ideal gas passing through the state having exactly \(p=1.0 \mathrm{~atm}\) and \(T=300 \mathrm{~K}\). Assume a diatomic gas whose molecules rotate but do not oscillate.

The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains. When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressure \(p\) depends on altitude \(y\) according to \(p=p_{0} \exp (-a y)\), where \(p_{0}=\) \(1.00\) atm and \(a=1.16 \times 10^{-4} \mathrm{~m}^{-1}\). Also assume that the ratio of the molar specific heats is \(\gamma=\frac{4}{3}\). A parcel of air with an initial temperature of \(-5.00^{\circ} \mathrm{C}\) descends adiabatically from \(y_{1}=4267 \mathrm{~m}\) to \(y=1567 \mathrm{~m}\). What is its temperature at the end of the descent?

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