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During the compression stroke of a certain gasoline engine, the pressure increases from 1.00 atm to 20.0 atm. If the process is adiabatic and the fuel- air mixture behaves as a diatomic ideal gas, (a) by what factor does the volume change and (b) by what factor does the temperature change? (c) Assuming that the compression starts with 0.0160 mol of gas at \(27.0^{\circ} \mathrm{C},\) find the values of \(Q, W,\) and \(\Delta E_{\text {int }}\) that characterize the process.

Short Answer

Expert verified
The volume changes by a factor calculated in Step 1. The temperature changes by a factor calculated in Step 2. The values of \(Q\), \(W\), and \(\Delta E_{\text {int }}\) are calculated in Step 3.

Step by step solution

01

Find the change in volume.

Use the known relationship for the adiabatic process of an ideal gas, \[PV^{\gamma} = constant\], where \(P\) is the pressure, \(V\) is the volume, and \(\gamma\) is the adiabatic index, which equals to 1.4 for a diatomic ideal gas. Therefore, \[P_1V_1^{\gamma } = P_2V_2^{\gamma }\]. Rearrange it to find the volume ratio \(V_1/V_2 = (P_2/P_1)^{1/\gamma}\), where \(P_1 = 1.00\, atm = 1.01 \times 10^{5}\, Pa\), \(P_2 = 20.0\, atm = 2.02 \times 10^{6}\, Pa\). Substitute these values and \(\gamma = 1.4\) to find \(V_1/V_2\).
02

Find the change in temperature.

The relationship between temperatures and volumes in an adiabatic process for ideal gases is \(T_1V_1^{\gamma -1} = T_2V_2^{\gamma -1}\). Therefore, \(T_1/T_2 = (V_2/V_1)^{\gamma - 1}\). We already know \(V_2/V_1\), so plug it in and use \(\gamma = 1.4\) to find \(T_1/T_2\).
03

Calculation of \(Q\), \(W\), and \(\Delta E_{\text {int }}\)

As the process is adiabatic, there is no heat exchange, which makes \(Q = 0\). Then, according to the first law of thermodynamics, \(W = \Delta E_{\text {int }} - Q\). To calculate \(\Delta E_{\text {int }}\), use the relationship \(\Delta E_{\text {int }} = C \Delta T\), where \(C\) is the heat capacity at constant volume equal to \(5/2R\), where \(R\) is the universal gas constant \(8.314\, J/(mol.K)\) and \(\Delta T = T_2 - T_1\). \(T_1 = 27.0^{\circ} \mathrm{C} = 300.15\, K \) and \(T_2 = T_1 \times (T_1/T_2)\) as calculated from the previous step. Hence by substituting these values, the change in internal energy \(\Delta E_{\text {int }}\) can be found. Finally, by substituting \(\Delta E_{\text {int }}\) and \(Q = 0\) into \(W = \Delta E_{\text {int }} - Q\), calculate \(W\).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry and physics that describes the behavior of an ideal gas under various conditions. This law combines three important gas laws: Boyle's Law, Charles's Law, and Avogadro's Law, into one comprehensive equation:
\[ PV = nRT \]
Where:
  • \( P \) is the pressure of the gas,
  • \( V \) is the volume,
  • \( n \) is the number of moles,
  • \( R \) is the universal gas constant (8.314 J/(mol·K)), and
  • \( T \) is the temperature in Kelvin.
Ideal gases perfectly follow this law without any exceptions, though real gases do deviate slightly. This formula is applied to calculate one parameter when the others are known.
In the context of an adiabatic process, the Ideal Gas Law helps connect the changes in pressure, volume, and temperature. When solving problems related to an adiabatic process, the relationship between pressure and volume becomes critical, and the adaptions derived from the Ideal Gas Law are used to solve for unknown variables.
Thermodynamics
Thermodynamics is the branch of physics studying heat, work, and the energy dynamics of systems. It is guided by four core laws, but the first and second laws are most pertinent here. The first law of thermodynamics states that energy can't be created or destroyed—only transformed from one form to another. In mathematical terms:
\[ \Delta E_{\text{int}} = Q - W \]
Where:
  • \( \Delta E_{\text{int}} \) is the change in internal energy,
  • \( Q \) is the heat absorbed or released, and
  • \( W \) is the work done by or on the system.
For adiabatic processes, there is no heat exchange, meaning \( Q = 0 \). Hence, the work done \( W \) is equivalent to the change in internal energy \( \Delta E_{\text{int}} \).
The second law of thermodynamics entails a system's eventual increase in entropy, guiding efficient energy conversion. Even in perfectly theoretical adiabatic processes, understanding thermodynamics is crucial for calculating the factors at play, such as changes in temperature, work, and internal energy.
Internal Energy Change
The internal energy change in a system, denoted \( \Delta E_{\text{int}} \), is critical for understanding energy dynamics in various processes, including adiabatic ones. It represents the total change in the energy contained within a system and measures how the energy distribution among particles changes due to temperature or state changes.
In an ideal gas, the internal energy change can be calculated using:
\[ \Delta E_{\text{int}} = C_v \Delta T \]
Where:
  • \( C_v \) is the molar heat capacity at constant volume -- for diatomic gases, it equals \( \frac{5}{2}R \).
  • \( \Delta T \) is the change in temperature.
During an adiabatic compression, because there is no heat transfer, the work done on or by the gas changes its internal energy. This change directly relates to the difference in the temperature of the system
(i.e., \( \Delta T = T_2 - T_1 \)). The value of \( \Delta E_{\text{int}} \) then helps determine other factors like the work, utilizing the first law of thermodynamics, further explaining how energy flow transforms within an isolated system. This thorough understanding, alongside the adiabatic principles, allows us to solve for changes in various parameters even under no heat exchange conditions.

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Most popular questions from this chapter

From the Maxwell-Boltzmann speed distribution, show that the most probable speed of a gas molecule is given by Equation \(21.29 .\) Note that the most probable speed corresponds to the point at which the slope of the speed distribution curve \(d N_{v} / d v\) is zero.

If you can't walk to outer space, can you at least walk halfway? Using the law of atmospheres from Problem \(43,\) we find that the average height of a molecule in the Earth's atmosphere is given by $$\bar{y}=\frac{\int_{0}^{\infty} y n_{V}(y) d y}{\int_{0}^{\infty} n_{V}(y) d y}=\frac{\int_{0}^{\infty} y e^{-m g / k_{B} T} d y}{\int_{0}^{\infty} e^{-m g / k_{B} T} d y}$$ (a) Prove that this average height is equal to \(k_{\mathrm{B}} T / m g\). (b) Evaluate the average height, assuming the temperature is \(10^{\circ} \mathrm{C}\) and the molecular mass is \(28.9 \mathrm{u}\)

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