During the time 0.332 mol of an ideal gas undergoes an isothermal compression at 26.4 c, 383 j of work is done on it by the surroundings. the final pressure is 1.87 atm. what was the intial pressure?

= 2.97 atm

Explanation:

The work in thermodynamic processes has several expressions, in the case of isothermal processes the work is

      W = n RT ln ( /  )

Where R is the constant of the ideal gases with a value of 8.314 J / mol K and T the absolute temperature

Let's look for the volume ratio

    T = 26.4 +273.15 = 299.55k

    Ln ( /  )

= W / n RT

    Ln( /  )

= 383 / (0.332 8.314 299.55)

    Ln( /  )

= 0.4632

     ( /  )  = e 0.4632

    ( /  )  = 1,589

Let's use the ideal gas equation for two volumes

     = n R T

    = n RT

The temperature is the same because it is an isothermal process, this equation is the same

      =

   Vi / Vf =  

  =     /  

    = 1.87 1,589

    = 2.97 atm