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Propane gas (\(C_3H_8\)) behaves like an ideal gas with \(_\Upsilon\) = 1.127. Determine the molar heat capacity at constant volume and the molar heat capacity at constant pressure.

Short Answer

Expert verified
\(C_v\) is approximately 65.45 J/(mol K) and \(C_p\) is approximately 73.76 J/(mol K).

Step by step solution

01

Identify Necessary Formulas

We need to determine two values: the molar heat capacity at constant volume, \(C_v\), and the molar heat capacity at constant pressure, \(C_p\). For an ideal gas, the relationship between \(C_v\) and \(C_p\) is given by the equation \(C_p = C_v + R\), where \(R\) is the universal gas constant, approximately 8.314 J/(mol·K).
02

Use Gamma Relation

The heat capacity ratio \(\gamma\) is defined as \(\gamma = \frac{C_p}{C_v}\). We're given \(\gamma = 1.127\). Using this formula allows us to write \(C_p = \gamma C_v\). We will use this to find the individual heat capacities.
03

Set Up Equations

Using the equations from the previous steps, we now have: 1. \(C_p = \gamma C_v\) 2. \(C_p = C_v + R\).
04

Solve for \(C_v\)

Substitute \(C_p = \gamma C_v\) from Step 3 into the equation \(C_p = C_v + R\): \(\gamma C_v = C_v + R\). This simplifies to \(C_v(\gamma - 1) = R\).
05

Calculate \(C_v\)

Solve for \(C_v\): \[ C_v = \frac{R}{\gamma - 1} = \frac{8.314}{1.127 - 1} \approx \frac{8.314}{0.127} \approx 65.45 \text{ J/(mol·K)}. \]
06

Calculate \(C_p\)

Now that we have \(C_v\), calculate \(C_p\) using \(C_p = C_v + R\): \[ C_p = 65.45 + 8.314 \approx 73.76 \text{ J/(mol·K)}. \]

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Key Concepts

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

Molar Heat Capacity
Molar heat capacity refers to the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius (or one Kelvin). In gases, this quantity is different depending on whether the pressure or volume is held constant during the heating process.
  • At constant volume (\(C_v\)): This is the heat capacity when no work is done by or on the gas because the volume of the gas does not change. Here, the energy goes entirely into raising the temperature.
  • At constant pressure (\(C_p\)): This is the heat capacity when the gas is allowed to expand, doing work on its surroundings. In this case, both the temperature and the volume of the gas change.
For propane gas, like many other ideal gases, the relationship between the two is given by \(C_p = C_v + R\), where \(R\) is the universal gas constant. This relationship underscores the fact that the molar heat capacity at constant pressure is always greater than at constant volume due to the work done during expansion.
Heat Capacity Ratio (Gamma)
The heat capacity ratio, often denoted as \(\gamma\), is a crucial parameter in thermodynamics, especially concerning gases. It is defined as the ratio of the molar heat capacities:\[ \gamma = \frac{C_p}{C_v} \]This ratio provides insight into how a gas will behave under adiabatic processes (where no heat is exchanged with the surroundings). In the given exercise, \(\gamma\) for propane is 1.127, reflecting its specific properties compared to other gases.
- **Adiabatic Process Insight**: Knowing the value of \(\gamma\) is vital as it affects the speed of sound in a gas and determines the pressure-volume relationship in adiabatic expansions.- **Finding \(C_v\) and \(C_p\)**: Given \(\gamma = 1.127\), it allows us to use both this relation and the equation \(C_p = C_v + R\) to solve for each heat capacity as shown in the step-by-step solution.
Propane Gas Characteristics
Propane (\(C_3H_8\)) is a commonly used hydrocarbon gas that demonstrates behavior similar to an ideal gas under many conditions. Here are some important characteristics of propane:- **Colorless and Odorless**: It's naturally colorless and odorless, but an odorant is usually added for safety reasons.- **Highly Flammable**: Propane is used extensively as a fuel source due to its high energy content.- **Applications**: It is commonly used for heating, cooking, and as a fuel for engines.When considering propane as an ideal gas in exercises, it follows the basic principles of the ideal gas law, allowing us to predict its behavior accurately using equations like the one in the given exercise involving \(C_p\), \(C_v\), \(\gamma\), and \(R\). Recognizing how these properties drive real-world applications helps understand why equations like the ideal gas law are so useful.

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Most popular questions from this chapter

A gas in a cylinder is held at a constant pressure of 1.80 \(\times\) 10\(^5\) \(Pa\) and is cooled and compressed from 1.70 m\(^3\) to 1.20 m\(^3\). The internal energy of the gas decreases by 1.40 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the absolute value of the heat flow, [\(Q\)] , into or out of the gas, and state the direction of the heat flow. (c) Does it matter whether the gas is ideal? Why or why not?

A cylinder contains 0.250 mol of carbon dioxide (\(CO_2\)) gas at a temperature of 27.0\(^\circ\)C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to 127.0\(^\circ\)C. Assume that the CO\(_2\) may be treated as an ideal gas. (a) Draw a \(pV\)-diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m\(^3\) and the pressure increases from 2.00 \(\times\) 10\(^5\) Pa to 5.00 \(\times\) 10\(^5\) Pa. The second process is a compression to a volume of 0.120 m\(^3\) at a constant pressure of 5.00 \(\times\) 10\(^5\) Pa. (a) In a pV-diagram, show both processes. (b) Find the total work done by the gas during both processes.

In another test, the valve of a 500-L cylinder full of the gas mixture at 2000 psi (gauge pressure) is opened wide so that the gas rushes out of the cylinder very rapidly. Why might some \(N_2O\) condense during this process? (a) This is an isochoric process in which the pressure decreases, so the temperature also decreases. (b) Because of the rapid expansion, heat is removed from the system, so the internal energy and temperature of the gas decrease. (c) This is an isobaric process, so as the volume increases, the temperature decreases proportionally. (d) With the rapid expansion, the expanding gas does work with no heat input, so the internal energy and temperature of the gas decrease.

A cylinder contains 0.100 mol of an ideal monatomic gas. Initially the gas is at 1.00 \(\times\) 10\(^5 \)Pa and occupies a volume of 2.50 \(\times\) 10\(^{-3}\) m\(^3\). (a) Find the initial temperature of the gas in kelvins. (b) If the gas is allowed to expand to twice the initial volume, find the final temperature (in kelvins) and pressure of the gas if the expansion is (i) isothermal; (ii) isobaric; (iii) adiabatic.

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