Calculating Work Done by Gas at Constant Pressure

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To calculate the work done by a gas at constant pressure, the formula W = P * ΔV is used, where W is work, P is pressure, and ΔV is the change in volume. In this scenario, the gas volume remains constant at 8m^3, so the work done is zero despite the temperature increase from 400K to 550K. The ideal gas law can provide insights into the behavior of the gas, but it is not necessary for calculating work in this case. Resources like Khan Academy and other educational sites can help clarify these concepts further. Understanding that work is related to volume change is crucial for solving similar problems.
pizzamakeren
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Homework Statement
How much work has the gas done?
Relevant Equations
K = Kelvin / Kpa = Kelvin Pascal
A gas with a volume of 8m^3 with a temperature of 400K gets warmed up to 550K with a constant pressure of 200Kpa. How much work has the gas done to the environment?

I think i might need to use the ideal gas law for this which is:
( P * V / T = N * K ) Where V is volume, T is temperature and N is the amount of atoms. P is pressure and K is a constant 1,38 * 10^-23J * K

The answer is supposed to be given in Kj, but I am unsure to how I am supposed to go forward.
 
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Are you familiar with the ideal gas law? Are you familiar with the equation for work in terms of pressure and volume change?
 
pizzamakeren said:
Homework Statement:: How much work has the gas done?
Relevant Equations:: K = Kelvin / Kpa = Kelvin Pascal

A gas with a volume of 8m^3 with a temperature of 400K gets warmed up to 550K with a constant pressure of 200Kpa. How much work has the gas done to the environment?

I think i might need to use the ideal gas law for this which is:
( P * V / T = N * K ) Where V is volume, T is temperature and N is the amount of atoms. P is pressure and K is a constant 1,38 * 10^-23J * K

The answer is supposed to be given in Kj, but I am unsure to how I am supposed to go forward.
I see that you've edited your Original Post after reading the hints/questions by @Chestermiller

So when the gas expands, it does work, right? A quick Google search provides some good learning resources if your textbook is not seeming to help...

https://www.ux1.eiu.edu/~cfadd/1150/14Thermo/work.html

https://www.khanacademy.org/science...ry/internal-energy-sal/a/pressure-volume-work
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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