Suppose 3 mols of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The gas is then allowed to expand quickly and adiabatically back to its original volume. Find the highest temperature attained by the gas. Find the lowest temperature attained by the gas. Find the highest pressure attained by the gas. Find the lowest pressure attained by the gas.

In an isothermal process, P*V = contant

In an adiabatic process (at constant entropy), P*V^gamma = constant
where gamma is the specific heat ratio, Cp/Cv. (Adiabatic expansions are isentropic unless they are into a vacuum and do no work)

For neon and other monatomic gases,
gamma = 5/3.

Original To = 273 K
Original Po = 1.00 atm
Original Vo = 3*22.4 = 67.2 liters
After the isothermal compression,
V1 = 12.77 liter = 0.19 Vo
T1 = 273 K
P1 = (1/0.19)Po = 5.26 Po
After the adiabatic expansion,
V2 = Vo
P2*V2^5/3 = P1*V1^5/3
P2/P1 = (V1/V2)^5/3 = (V1/V0)^5/3
= 0.19^5/3 = 0.0628
P2 = 0.0628 *5.26 = 0.330 atm
P2*V2/T2 = Po*Vo/To
T2 = (P2/Po)(V2/Vo)*To
= 0.330* 273 = 90.2 K

Highest P = Po = 5.26 atm
Lowest P = P2 = 0.330 atm
Highest T = To
Lowest T = T2

Well, well, well, it seems like we have a real gas party going on here! Let's see what's cooking.

First things first, let's tackle the highest temperature attained by our gas. Since the compression is isothermal, we can use the ideal gas law to find the initial temperature. Tada! We get T₁ = P₁V₁ / (n₁R), where P₁ is the initial pressure, V₁ is the initial volume, n₁ is the initial number of moles, and R is the ideal gas constant.

Now, during the adiabatic expansion, the temperature and volume relationship is governed by the equation PV^(γ) = constant, where γ is the heat capacity ratio (specific heat capacity at constant pressure divided by the specific heat capacity at constant volume). In this case, since neon is a monatomic ideal gas, γ = 5/3.

Using that equation, we find that T₂ = T₁ (V₁/V₂)^((γ-1)/γ), where T₂ is the final temperature and V₂ is the final volume.

For the lowest temperature, we know that the adiabatic expansion is a rapid process (no heat exchange), so the lowest temperature is simply the final temperature T₂.

As for the highest pressure reached by our gas, we can once again rely on that adiabatic expansion equation. Using PV^(γ) = constant, we find that P₂ = P₁ (V₁/V₂)^γ.

The lowest pressure attained by our gas is simply the final pressure P₂.

Now, I could go through all the calculations step by step, but hey, where's the fun in that? Let's grab our handy-dandy calculator or nerd glasses and apply these formulas to find the specific answers. Enjoy the gas ride!

To solve this problem, we can use the ideal gas law and the adiabatic expansion/compression equations.

Step 1: Find the initial conditions at STP.
At STP (Standard Temperature and Pressure), the conditions are:
Temperature (T1) = 273 K
Pressure (P1) = 1 atm
Volume (V1) = 22.4 L (since 1 mole of any ideal gas occupies 22.4 L at STP)

Step 2: Compression (isothermal process)
The gas is compressed slowly and isothermally to 0.19 times its original volume.
Volume (V2) = 0.19 * V1 = 0.19 * 22.4 L = 4.256 L

Since the process is isothermal, the temperature remains constant at T1, which is 273 K.

Using the ideal gas law, we can find the new pressure (P2):
P1 * V1 = P2 * V2
1 atm * 22.4 L = P2 * 4.256 L
P2 = (1 atm * 22.4 L) / 4.256 L
P2 = 5.27 atm (rounded to two decimal places)

Step 3: Adiabatic expansion (quick and without heat transfer)
The gas expands quickly and adiabatically back to its original volume, V1.

Using the adiabatic expansion equation:
P1 * V1^(γ) = P2 * V2^(γ),
where γ is the heat capacity ratio (Cp/Cv) for a monatomic ideal gas, which is equal to 5/3.

Substituting the known values:
1 atm * (22.4 L)^(5/3) = 5.27 atm * (4.256 L)^(5/3)

Now, we can solve for the final pressure attained during expansion, P3:

P3 = (1 atm * (22.4 L)^(5/3)) / (4.256 L)^(5/3)
P3 = 0.19 atm (rounded to two decimal places)

Step 4: Find the highest and lowest temperatures attained during the process.

Highest temperature: Since the process is adiabatic, there is no heat transfer, and the highest temperature will occur during the compression process. The temperature remains constant at T1, which is 273 K.

Lowest temperature: The lowest temperature will occur during the expansion process, which is also adiabatic. We can use the adiabatic expansion equation to find the lowest temperature, T3:

T1 * (P1 / P3)^((γ-1)/γ) = T3 * (P3 / P3)^((γ-1)/γ)
273 K * (1 atm / 0.19 atm)^(2/5) = T3 * 1
T3 = 393.3 K (rounded to one decimal place)

Step 5: Summary of the results:

Highest temperature attained by the gas: 273 K
Lowest temperature attained by the gas: 393.3 K
Highest pressure attained by the gas: 5.27 atm
Lowest pressure attained by the gas: 0.19 atm

To solve this problem, we need to use the ideal gas law and the relationships between the variables during a compression or expansion process.

1. Let's start by finding the highest temperature attained by the gas during the expansion process. Since the expansion is adiabatic, it means no heat is exchanged with the surroundings. We can use the adiabatic expansion formula:

(T2 / T1) = (V1 / V2) ^ (γ - 1)

where T2 is the final temperature, T1 is the initial temperature, V1 is the initial volume, V2 is the final volume, and γ is the heat capacity ratio for a monatomic gas, which is 5/3.

Given that the volume expands back to its original value (V2 = V1), and we know that the initial temperature (T1) is 273.15 K (STP - Standard Temperature and Pressure), we can rearrange the equation to solve for T2:

(T2 / 273.15K) = (1 / 0.19) ^ (5/3 - 1)

T2 = (1 / 0.19) ^ (5/3 - 1) * 273.15K

Calculate the value to find the highest temperature attained by the gas during the expansion.

2. Next, let's find the lowest temperature attained by the gas during the compression process. Since the compression is isothermal, it means the temperature remains constant. Therefore, the lowest temperature attained by the gas during the compression process will be the initial temperature, which is 273.15 K.

3. To find the highest pressure attained by the gas during the compression process, we can again use the ideal gas law. Since the process is slow and isothermal, the temperature remains constant at 273.15 K. Therefore, the equation can be written as:

P1 * V1 = P2 * V2

where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

Given that the final volume (V2) is 0.19 times the initial volume (V1) and the initial pressure (P1) is calculated using the ideal gas law for STP (P1 = nRT1/V1), we can rearrange the equation to solve for P2:

P2 = P1 * V1 / V2

Calculate the value to find the highest pressure attained by the gas during the compression.

4. Finally, to find the lowest pressure attained by the gas during the expansion process, we can use the relationship between pressure and volume during an adiabatic process:

P2 / P1 = (V1 / V2) ^ γ

Where P2 is the final pressure, P1 is the initial pressure, V1 is the initial volume, V2 is the final volume, and γ is the heat capacity ratio (5/3).

Given that the final volume (V2) is the same as the initial volume (V1), and the initial pressure (P1) is calculated using the ideal gas law for STP (P1 = nRT1/V1), we can rearrange the equation to solve for P2:

P2 = P1 * (V1 / V2) ^ γ

Calculate the value to find the lowest pressure attained by the gas during the expansion.

Solving these equations will provide the answers to the highest and lowest temperatures, as well as the highest and lowest pressures attained by the gas during the given processes.