Question

Propane gas ($C_3{H}_8$) behaves like an ideal gas with ${}_{\mathrm{{\rm Y}}}$ = 1.127. Determine the molar heat capacity at constant volume and the molar heat capacity at constant pressure.

Short answer

$C_v$ is approximately 65.45 J/(mol K) and $C_p$ is approximately 73.76 J/(mol K).

Step-by-step solution

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).

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.

Set Up Equations

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

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 $C_v(\gamma -1)=R$.

Calculate $C_v$

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

Calculate $C_p$

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

Key concepts

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 ($C_3{H}_8$) 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.