# Work under constant temperature

• ymehuuh
In summary, an ideal monatomic gas expands isothermally from 0.590 m3 to 1.25 m3 at a constant temperature of 780 K. To find the work done on the gas, we can use the equation W= -PV. The relationship between P1V1 and P2V2 is given by PV=nRT, and using this we can find the final pressure, P2.

## Homework Statement

An ideal monatomic gas expands isothermally from 0.590 m3 to 1.25 m3 at a constant temperature of 780 K. If the initial pressure is 1.20x10^5 Pa.
(a) Find the work done on the gas.
(b) Find the thermal energy transfer Q.
(c) Find the change in the internal energy.

## Homework Equations

W= -PV
deltaU=Q-W
PfVf/PiVi = Tf/Ti

## The Attempt at a Solution

I know I need to find the final pressure...but I'm not sure how to.
Do I use Pf*1.35/1.2x10^5 = 780k?

ymehuuh said:

## Homework Statement

An ideal monatomic gas expands isothermally from 0.590 m3 to 1.25 m3 at a constant temperature of 780 K. If the initial pressure is 1.20x10^5 Pa.
(a) Find the work done on the gas.
If T is constant and PV=nRT, what is the relationship between P1V1 and P2V2? Does that help you to find P2?

AM

I would approach this problem by first acknowledging that the work is being done under constant temperature, meaning that the temperature of the gas remains constant throughout the expansion process. This implies that the ideal gas law (PV = nRT) can be simplified to P1V1 = P2V2, where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume.

(a) To find the work done on the gas, we can use the equation W = -PΔV, where ΔV is the change in volume. In this case, ΔV = V2 - V1 = 1.25 m^3 - 0.590 m^3 = 0.660 m^3. Therefore, the work done on the gas is W = -(1.20x10^5 Pa)(0.660 m^3) = -79,200 J.

(b) To find the thermal energy transfer Q, we can use the first law of thermodynamics, which states that ΔU = Q - W, where ΔU is the change in internal energy. Since the temperature remains constant, ΔU = 0 and thus Q = W. Therefore, Q = 79,200 J.

(c) To find the change in internal energy, we can use the ideal gas law to calculate the final pressure, P2 = P1V1/V2 = (1.20x10^5 Pa)(0.590 m^3)/(1.25 m^3) = 56,640 Pa. We can then use the equation ΔU = nCvΔT, where n is the number of moles of gas, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature. Since the temperature remains constant, ΔT = 0 and thus ΔU = 0. This makes sense because the internal energy of an ideal gas only depends on its temperature, which remains constant in this case.

## 1. What is constant temperature work?

Constant temperature work refers to any type of work or experiment that is conducted at a specific and unchanging temperature. This is often done to maintain consistent and controlled conditions in order to obtain accurate and repeatable results.

## 2. Why is it important to work under constant temperature?

Working under constant temperature is important because many chemical and biological reactions are highly sensitive to temperature. Small changes in temperature can significantly affect the outcome of an experiment, so by maintaining a constant temperature, scientists can eliminate this variable and ensure more accurate results.

## 3. What techniques are used for working under constant temperature?

There are several techniques that can be used for working under constant temperature, including using a water bath, heating mantle, or temperature-controlled incubator. These tools can help maintain a consistent temperature throughout an experiment.

## 4. What are the benefits of working under constant temperature?

The benefits of working under constant temperature include increased accuracy and reproducibility of results, as well as the ability to control and monitor reactions more closely. This can also save time and resources by reducing the need for repeated experiments due to temperature fluctuations.

## 5. Are there any limitations to working under constant temperature?

One limitation of working under constant temperature is that it may not accurately reflect real-world conditions, where temperature is constantly changing. Additionally, some experiments may require varying temperatures in order to observe certain reactions or processes. In these cases, working under constant temperature may not be suitable.

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