# Calculating Internal Energy and Pressure Volume Work in an Argon Balloon

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## Introduction

In this article, we’ll explore a practical example of calculating the internal energy and pressure-volume work for an argon gas balloon. We will navigate through the concepts of the first law of thermodynamics, analyze the energy transfer, and ensure correct unit conversions throughout the process.

## Understanding Thermodynamics and the First Law

Thermodynamics is the study of heat, energy, and work, and the first law of thermodynamics states that energy cannot be created or destroyed, only transformed. For any thermodynamic process, the change in internal energy (ΔU) of a system can be determined by the equation:

[ ΔU = Q + W ] Where:

**Q**= Heat added to the system (positive) or released from the system (negative)**W**= Work done on the system (positive) or by the system (negative)

### Problem Setup

In our example, we consider:

**External Pressure (P)**: 1.01 × 10^5 Pascal**Initial Volume (V1)**: 2.3 liters**Final Volume (V2)**: 2.05 liters**Heat transferred to surroundings (Q)**: –485 joules

Given these values, we are tasked with calculating the change in internal energy (ΔU) for this process.

## Step-by-Step Calculation

To simplify the calculations, we will follow these steps:

### 1. Assign Signs to Heat and Work

Before plugging in any values, it’s crucial to assign appropriate signs:

- Since our system (the balloon) transfers energy to the surroundings,
**Q**is negative:**Q = –485 J**. - The volume decreases from V1 to V2, indicating the surroundings did work on our system. Thus,
**W**is positive.

### 2. Calculate Work Done (W)

The work done on the gas can be calculated using the formula for pressure-volume work: [ W = P × (V2 - V1) ]

- Where
**V2 - V1**= Final Volume - Initial Volume = 2.05 L – 2.3 L = -0.25 L - We need to convert liters to cubic meters (1 L = 0.001 m³):

[ W = 1.01 × 10^5 ext{ Pa} × (-0.25 ext{ L}) = 1.01 × 10^5 ext{ Pa} × (-0.00025 ext{ m}³) ] - Calculating gives us:
[ W = –25.25 ext{ J} ]

Since work is done**on**the system, it should be positive. Therefore, we notice that the overall work was performed**by**the system.

### 3. Plugging Values into the First Law

Now that we have determined our values, let’s substitute them into the first law equation:
[ ΔU = Q + W ]

Substituting in the calculated values:

- Q = –485 J
- W = –25.25 J

[ ΔU = -485 ext{ J} + (-25.25 ext{ J}) ]

Performing this calculation results in:
[ ΔU = -510.25 ext{ J} ]

(Note: Unifying our work signs is crucial in this calculation process.)

### 4. Verifying Units

An essential step in our calculations is ensuring that our units make sense. In terms of SI units:

- 1 Joule (J) = 1 kg·m²/s²
- Pascal x Volume (L to m³) also needs conversion to ensure both sides of the equation are in Joules.

This consistency reassures us that our calculations maintain accuracy.

## Conclusion

In conclusion, calculating internal energy and pressure-volume work is central to understanding thermodynamic processes in physical systems. From our example, we have derived:

- The change in internal energy (ΔU) of the argon gas balloon is
**–510.25 J**. - Key takeaways include always noting the signs for heat and work and confirming unit consistency throughout calculations. Practicing these principles will enhance your understanding and efficiency in solving thermodynamic problems.

in this video we're going to do an example problem where we calculate internal energy and also calculate

pressure volume work so we know the external pressure is 1.01 * 10 5th Pascal and our system is

some balloon let's say it's a balloon of argon gas and initially our gas has a volume

what is Delta U so what's a change in internal energy for our system we can use a first law of thermodynamics to

answer this the first law tells us that the change in internal energy Delta U is equal to the work

done plus the heat transfer before we plug any numbers in here the first thing I want to do is make sure I have a good

idea of what signs everything should be I think that's one of the trickiest things things and these kind of problems

so here since our system transferred energy to the surroundings and not the other way around Q should be negative

because when it your system transfers energy to the surroundings then its internal energy should go down work on

the other hand since V2 is less than V1 the volume of our system went down which means the surroundings had to do work on

the gas to get the volume to decrease we would expect if the surroundings did work on our system that would increase

the internal energy so that means the work done here is positive we can also calculate work because we know the

external pressure we know it's constant and work can be calculated as the external pressure times the change in

volume and we know both of those things we know the external pressure and we know the initial and final volumes so if

we start plugging that in we get that Delta U is equal to -485 so that's our heat we know it

should have a negative sign because the heat was transferred to the surroundings so -485 Jew

pascals so that's our external pressure times our change in volume so that's that's our final pressure 2.05 L minus

our initial pressure which is 2.30 L we could at this point be like okay we figured it all out we just have

to stick all of these numbers in the calculator and we're done and that's probably what my first instinct would

often be but there there's one more thing that we should check before we actually plug in the numbers and and

have a party and that's our units we have our heat in terms of jewels we probably want our change in internal

energy in terms of jewels too on this other side we're calculating our work here and we have pascals so pascals

times so we have jewels and jewels we have pascals time liters so then the question is okay we're doing Jewels

minus pascals time liters we need to make sure that whatever we calculate here in terms of work also has units of

jewels otherwise we will be will be subtracting two things that don't have the same units and that's bad we'll have

to do some sort of unit conversion first so let's just double check that pascals times lers will give us Jewels so the

way I did this is by converting everything to the same units so if you take Jewels which is

already s units we can actually simplify it more in terms of other essay units so a jewel is equal to 1 kog M squared per

pascal pascals are also in terms of are also SI units and if we convert pascals to kilogram meters and seconds we get

multiply this by units of volume and whatever we multiply it by should give us units of jewels so what we need to do

here is convert our lers to me cubed and if we do that everything is in terms of kilogram meters and seconds

this meters cancels out with one of these and we end up with 1 kilg m squ per second squared on both sides so then

everything is in terms of jewels that's not the only way you could have made sure that the units made sense

um you could have converted them to something else so basically you just have to make sure that all of the units

you're using in your equation match each other if you're going to add them or subtract them all of this is to say that

we need to make sure we convert liters to meters cubed so that everything works out in terms of jewels so if we do that

Pascal * .25 L which is a change in volume which is negative the volume went down so the change in volume should be

so now our liters cancel out and pascals time me cubed gives us Jews so that gives us that Delta U our change in

internal energy is -485 Jew then if we plug this all into our calculator to calculate the work we get positive

Jew so the key things to remember here for this kind of problem is to double check your signs for work and heat and