Argon Gas Expansion Work And Heat Released Calculation In Chemistry

Hey guys! Ever wondered how to calculate the work done when a gas expands or the heat released when something cools down? Well, today we're diving into two super interesting chemistry problems. We'll break down how to figure out the work done by expanding argon gas and how much heat is released when water cools. Let's get started!

1. Work Done by Expanding Argon Gas

Let's tackle the first part of our challenge: calculating the work done when a sample of argon gas expands. This involves understanding the conditions under which the expansion occurs, specifically against a vacuum and against a constant pressure.

Understanding the Basics of Work in Thermodynamics

In thermodynamics, work is the energy transferred when a force causes displacement. For gases, this often involves expansion or compression. The formula we use to calculate work (*W*) done by a gas is:

${ W = -P \Delta V }$

Where:

• ${ P }$ is the external pressure against which the gas expands.
• ${ \Delta V }$ is the change in volume (final volume minus initial volume).

It’s important to note the negative sign, which indicates that work done by the system (like the expanding gas) is considered negative, as the system loses energy.

(a) Expansion Against a Vacuum

First, let's consider the scenario where argon gas expands against a vacuum. A vacuum, by definition, has zero pressure. This simplifies our calculation significantly. If the external pressure (*P*) is 0 atm, then the work done (*W*) is:

${ W = -0 \times \Delta V = 0 \text{ joules} }$

So, when a gas expands into a vacuum, no work is done. This might seem counterintuitive, but it makes sense because there's no external force resisting the expansion. The gas is essentially expanding into empty space, encountering no opposition.

(b) Expansion Against a Constant Pressure of 4.2 atm

Now, let’s look at the more common scenario where the gas expands against a constant external pressure. In this case, the argon gas expands from an initial volume of 2.0 L to a final volume of 6.2 L against a constant pressure of 4.2 atm. To calculate the work done, we use the same formula:

${ W = -P \Delta V }$

First, we need to calculate the change in volume (${ \Delta V }$): ${ \Delta V = V_{\text{final}} - V_{\text{initial}} = 6.2 \text{ L} - 2.0 \text{ L} = 4.2 \text{ L} }$

Now we can plug the values into our formula:

${ W = -(4.2 \text{ atm}) \times (4.2 \text{ L}) = -17.64 \text{ L atm} }$

However, we need the work in joules, not liter-atmospheres. To convert L atm to joules, we use the conversion factor:

${ 1 \text{ L atm} = 101.325 \text{ J} }$

So,

${ W = -17.64 \text{ L atm} \times 101.325 \frac{\text{J}}{\text{L atm}} \approx -1787.4 \text{ J} }$

Therefore, the work done when the argon gas expands against a constant pressure of 4.2 atm is approximately -1787.4 joules. The negative sign indicates that the gas is doing work on the surroundings.

Key Takeaways for Gas Expansion Work

• Work done against a vacuum is always zero because there's no opposing pressure.
• Work done against a constant pressure involves calculating the change in volume and converting units if necessary.
• The negative sign in the work calculation indicates that the system (the gas) is doing work on the surroundings.

2. Heat Released When Water Cools

Moving on to our second problem, we need to determine the amount of heat released when 200 g of water cools from 85°C to 22°C. This involves the concept of specific heat capacity.

Understanding Specific Heat Capacity

Specific heat capacity (*c*) is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 Kelvin). Water has a relatively high specific heat capacity, which is approximately 4.184 J/g°C. This means it takes a significant amount of energy to change the temperature of water.

The formula to calculate the heat (*q*) released or absorbed is:

${ q = mc \Delta T }$

Where:

• ${ m }$ is the mass of the substance.
• ${ c }$ is the specific heat capacity of the substance.
• ${ \Delta T }$ is the change in temperature (final temperature minus initial temperature).

Calculating Heat Released by Cooling Water

In our case:

• Mass (*m*) = 200 g
• Specific heat capacity (*c*) = 4.184 J/g°C
• Initial temperature (*T_{\text{initial}}*) = 85°C
• Final temperature (*T_{\text{final}}*) = 22°C

First, we calculate the change in temperature (${ \Delta T }$):

${ \Delta T = T_{\text{final}} - T_{\text{initial}} = 22°C - 85°C = -63°C }$

Now, we can plug the values into our formula:

${ q = (200 \text{ g}) \times (4.184 \frac{\text{J}}{\text{g°C}}) \times (-63°C) \approx -52710.24 \text{ J} }$

So, the heat released when 200 g of water cools from 85°C to 22°C is approximately -52710.24 joules. The negative sign indicates that heat is being released by the water (an exothermic process).

To express this in kilojoules (kJ), we divide by 1000:

${ q \approx -52.71 \text{ kJ} }$

Key Points for Heat Release Calculation

• Specific heat capacity is crucial for determining how much heat is needed to change a substance's temperature.
• The formula ${ q = mc \Delta T }$ is fundamental for heat calculations.
• A negative value for heat (*q*) indicates heat is released (exothermic), while a positive value indicates heat is absorbed (endothermic).

Conclusion: Mastering Thermodynamics Calculations

Alright, guys, we've tackled two fascinating chemistry problems today! We calculated the work done by expanding argon gas under different conditions and determined the heat released when water cools. Understanding these concepts is essential for anyone diving into thermodynamics.

Remember, the work done by a gas depends on the external pressure, and the heat released or absorbed depends on the specific heat capacity of the substance. Keep these principles in mind, and you'll be solving similar problems like a pro in no time!

If you have any questions or want to explore more chemistry topics, drop a comment below. Keep learning, and I'll catch you in the next one!