Question

Gas occupies a volume of \(4.33 \mathrm{~L}\) at a pressure of \(1.17 \mathrm{~atm}\) and a temperature of \(310 \mathrm{~K}\). It is compressed adiabatically to a volume of \(1.06 \mathrm{~L}\). Determine \((a)\) the final pressure and \((b)\) the final temperature, assuming the gas to be an ideal gas for which \(\gamma=1.40 .(c)\) How much work was done on the gas?

Short answer

The final pressure, final temperature, and work done on the gas will be calculated based on the given initial conditions and the formulas applicable to an adiabatic process.

Step-by-step solution

Determine the final pressure according to the adiabatic process formula

We are given the initial pressure \(P_1 = 1.17 \) atm, the initial volume \(V_1 = 4.33 \) L, the final volume \(V_2 = 1.06 \) L, and the specific heat ratio \(\gamma = 1.4\). We can substitute those values into the adiabatic process formula \(P_1V_1^{\gamma} = P_2V_2^{\gamma}\) and solve it for the final pressure \(P_2\). This gives \(P_2 = P_1V_1^{\gamma} / V_2^{\gamma}\).

Determine the final temperature of the gas.

The ideal gas law equation can assist in obtaining the final temperature of the gas. The equation is \(P_1V_1 /T_1 = P_2V_2 / T_2\) where \(P\) is the pressure, \(V\) is the volume and \(T\) is the temperature. We know the values for \(P_1\), \(V_1\), \(T_1\), \(P_2\), and \(V_2\). Therefore, we can solve the equation for the final temperature, \(T_2\). This gives \(T_2 = P_2V_2T_1 / (P_1V_1)\).

Calculate the work done on the gas.

The work done (\(W\)) can be computed using the formula \(W = (P_2V_2 - P_1V_1) / (1 - \gamma)\). Here, we can input the final pressure, \(P_2\) and volume, \(V_2\), that we computed in Steps 1 and 2, along with the initial pressure, \(P_1\) and volume, \(V_1\), and the specific heat ratio, \(\gamma\), which was given, and compute the amount of work done on the gas.

Key concepts

ideal gas law

The ideal gas law is central to understanding how gases behave under various conditions. It combines several gas laws into one comprehensive equation: \( PV = nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. In our scenario, we are asked to find the final temperature of a gas that undergoes an adiabatic process. The ideal gas law helps us determine the relationship between pressure, volume, and temperature. Through the use of the equation \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \), we can solve for the unknown final temperature \( T_2 \).
This gives us a way to understand how, even without heat exchange in an adiabatic process, the energy transfer via work affects the system's temperature.

specific heat ratio

The specific heat ratio, denoted by \( \gamma \), is a crucial parameter in thermodynamics, especially when working with adiabatic processes. It is defined as the ratio of the specific heat at constant pressure \( C_p \) to the specific heat at constant volume \( C_v \).

• Mathematically: \( \gamma = \frac{C_p}{C_v} \)
• For an ideal gas, \( \gamma \) tells us how the pressure and volume of the gas are interrelated when there is no heat transfer, like in an adiabatic process.

In the problem at hand, we used a specific heat ratio of \( 1.4 \), which is typical for diatomic gases like oxygen and nitrogen under normal conditions. This ratio helps us calculate changes in other properties of the gas, such as pressure changes during the adiabatic compression using the formula \( P_1V_1^{\gamma} = P_2V_2^{\gamma} \). This neatly ties together the relationship between pressure and volume changes in scenarios without heat exchange.

work done on gas

When a gas is compressed adiabatically, work is done on the gas to decrease its volume and increase its pressure, as no energy is exchanged as heat. The work done can be calculated using the specific formula for adiabatic processes: \( W = \frac{P_2V_2 - P_1V_1}{1 - \gamma} \).
This formula is derived considering both the initial and final states of the gas, taking into account its pressure and volume.
The term \( \gamma \) in the denominator reflects how the specific heat ratio affects the amount of work done.
Calculating the work done provides insights into the energy dynamics involved in compressing the gas.

• High values of work occur if large changes in volume and pressure are achieved.
• Understanding this concept is key in applications like engines and refrigeration, where efficiency often depends on minimizing or controlling the work done on or by the gas.

Such calculations give us not only the amount of energy transferred but also highlight the intrinsic properties that define the gas's behavior under compression.