Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \(C_p = C_v + R\): \[ C_p = 65.45 + 8.314 \approx 73.76 \text{ J/(mol·K)}. \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two moles of an ideal monatomic gas go through the cycle \(abc\). For the complete cycle, 800 J of heat flows out of the gas. Process \(ab\) is at constant pressure, and process \(bc\) is at constant volume. States \(a\) and \(b\) have temperatures \(T_a\) = 200 K and \(T_b\) = 300 K. (a) Sketch the \(pV\)-diagram for the cycle. (b) What is the work \(W\) for the process \(ca\)?

The engine of a Ferrari F355 F1 sports car takes in air at 20.0\(^\circ\)C and 1.00 atm and compresses it adiabatically to 0.0900 times the original volume. The air may be treated as an ideal gas with \(_\Upsilon\) = 1.40. (a) Draw a \(pV\)-diagram for this process. (b) Find the final temperature and pressure.

A large research balloon containing \(2.00 \times 10^{3} \mathrm{~m}^{3}\) of helium gas at 1.00 atm and a temperature of \(15.0^{\circ} \mathrm{C}\) rises rapidly from ground level to an altitude at which the atmospheric pressure is only 0.900 atm (Fig. \(\mathbf{P} 19.50\) ). Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. (a) Calculate the volume of the gas at the higher altitude. (b) Calculate the temperature of the gas at the higher altitude. (c) What is the change in internal energy of the helium as the balloon rises to the higher altitude?

A player bounces a basketball on the floor, compressing it to 80.0% of its original volume. The air (assume it is essentially N\(_2\) gas) inside the ball is originally at 20.0\(^\circ\)C and 2.00 atm. The ball's inside diameter is 23.9 cm. (a) What temperature does the air in the ball reach at its maximum compression? Assume the compression is adiabatic and treat the gas as ideal. (b) By how much does the internal energy of the air change between the ball's original state and its maximum compression?

During the time 0.305 mol of an ideal gas undergoes an isothermal compression at 22.0\(^\circ\)C, 392 J of work is done on it by the surroundings. (a) If the final pressure is 1.76 atm, what was the initial pressure? (b) Sketch a \(pV\)-diagram for the process.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free