# How much work is done by the gas?

• Tehy
In summary, the conversation is discussing how to solve an exercise involving 5 moles of gas at a temperature of 326K and a pressure that is expanded from 1atm to 3atm. The question asks how much work is done by the gas. The expert provides a detailed explanation, taking into consideration external pressure changes and heat being added or released. They also mention that if the temperature is kept constant, the work done can be calculated using the formula W = -nRTln(Vi/Vf). However, if the temperature increases due to adiabatic conditions, the calculation is more complicated.
Tehy
Hello! Any ideas how I could solve this exercise:

5 mol of gas is in temperature of 326K and the same time the pressure is expanded from 1atm to 3atm. How much work is done by the gas?

Detailed instruction would be good, thanks :)

W/mole = dPV = Pdv + Vdp.

it should be pretty obvious from here.

sicjeff said:
W/mole = dPV = Pdv + Vdp.

it should be pretty obvious from here.

Yes, it probably should, but I'm very dumb with physics :P I already tried to find some help from Google but I didn't have any luck :/

okay, I assume you are in like a freshman/sophmore level physics course right.

Let's assume an ideal gas.

we know the standard molar volume right? (22.414 L/mole). so v=nRT/P.
We are assuming that the size of our container is constant. (isochoric condition)

sicjeff said:
okay, I assume you are in like a freshman/sophmore level physics course right.

Let's assume an ideal gas.

we know the standard molar volume right? (22.414 L/mole). so v=nRT/P.
We are assuming that the size of our container is constant. (isochoric condition)

Oh, yes it's ideal gas. I forget to mention that :)

Do I need to calculate v1= nRT/1atm and then v2= nRT/3atm?

Tehy said:
Hello! Any ideas how I could solve this exercise:

5 mol of gas is in temperature of 326K and the same time the pressure is expanded from 1atm to 3atm. How much work is done by the gas?

Detailed instruction would be good, thanks :)
The question does not have sufficient information. You have to know how the pressure is changed and whether heat is being added or released.

If the external pressure is increased, the volume must decrease, in which case the work done by the gas is negative. If pressure is increased because heat is added, the volume can be held constant, in which case, no work is done. Or it could increase, in which case the work done by the gas is positive.

Work = Pdv not $\Delta (PV)$

If you assume that the external pressure is gradually changed from 1 atm to 3 atm, and the temperature is kept constant, the work done is:

$$W = \int_{P_i}^{P_f} PdV = \int_{P_i}^{P_f} nRTdV/V = nRT\ln(\frac{V_f}{V_i}) = -nRT\ln(\frac{V_i}{V_f})$$

If it is adiabatic (temperature will increase) it is more complicated.

AM

Thanks Andrew! That works great! :)

## 1. How do you calculate the work done by a gas?

The work done by a gas can be calculated using the formula: W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume of the gas.

## 2. What is the unit of measurement for work done by a gas?

The unit of measurement for work done by a gas is joules (J). However, in some cases, it may also be measured in other units such as calories or kilogram-force meters (kgf·m).

## 3. Does the work done by a gas depend on the direction of expansion or compression?

Yes, the work done by a gas is a signed quantity and depends on the direction of expansion or compression. If the gas expands, the work done will be positive, whereas if the gas is compressed, the work done will be negative.

## 4. Can the work done by a gas be zero?

Yes, the work done by a gas can be zero if there is no change in volume or pressure. This can happen when the gas is in a state of equilibrium, where the external pressure is equal to the internal pressure of the gas.

## 5. How does the work done by a gas relate to its internal energy?

The work done by a gas is directly related to its internal energy. When work is done on a gas, its internal energy increases, and when work is done by a gas, its internal energy decreases. This is known as the first law of thermodynamics.

• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
661
• Introductory Physics Homework Help
Replies
11
Views
4K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
474
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
5K
• Introductory Physics Homework Help
Replies
21
Views
2K