How Does Isothermal Expansion Affect Work Done and Temperature in an Ideal Gas?

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In an isothermal expansion of 145 moles of a monatomic ideal gas from 1.00 m³ to 4.00 m³, the temperature remains constant throughout the process. The ideal gas law (PV=nRT) can be used to determine the temperature by substituting the initial pressure and volume values. The work done by the gas during the expansion can be calculated using the formula W=nRTln(vf/vi). The calculated temperature is approximately 0.332 K, and the work done is about 1670.42 J. The discussion confirms that the temperature does not change during isothermal expansion.
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Suppose 145 moles of a monatomic ideal gas undergo an isothermal expansion from 1.00m^3 to 4.00m^3, as shown in figure. what is the temp at the beginning and end of this process? how much work is done by the gas during the expansion?

http://www.physics.brocku.ca/Courses/1p23/images/FG18_24.JPG

since it is isothermal the temperature is constant correct? :confused:
so to solve for the temperature would i just use pv=nrt with the respective values for pressure and volume?
and for the work done W=nrtln(vf/vi)

thanks
 
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You have p_i and p_f, and also v_i and v_f. Can you think of a formula that uses p and v ?

Hint: What are the units for pressure?
 
so i wouldn't use pv=nrt?
 
Try it out. Plug in the values of p_i and v_i into that formula.
Then plug in the values of p_f and v_f into that formula.
See what you get.
 
im getting .332 for the temperature.
 
and for the work i get 1670.42j
 
Now you tell me. Is temperature constant?
 
yup yup. thought it was tryin to trick me by askin the temp of beginnin and end like it was going to be two different values.
 

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