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a sample of an ideal gas in a cylinder of volume 4.42 l at 298 k and 2.36 atm expands to 7.69 l by two different pathways. path a is an isothermal, reversible expansion. path b has two steps. in the first step, the gas is cooled at constant volume to 1.31 atm . in the second step, the gas is heated and allowed to expand against a constant external pressure of 1.31 atm until the final volume is 7.69 l . calculate the work for path a.

Respuesta :

Given in question;
Pressure (P) - 2.36atm
Volume of gas (V) - 4.42L
Temperature (T) - 298K

From the ideal gas law [pv = nrt]
; where P is pressure of gas
             V is volume of gas
             n is number of moles
                         R is gas constant
             T is temperature of gas

No of moles in Ideal solution(n) = RT/PV
Thus,
         n = (0.0821 atm/mol•K)(298K) / (2.26atm)(4.42L) …..{here R= 0.0821 atm/mol•K}
         n = 2.34 mol

To get Work done by path A,we will use
       W = -nRT ln(V¹/V²)
       {Where V¹ and V² is given 7.692 L and 4.422 L respectively}
thus,
       W=(2.34mol)(8.314 J/mol•K)(298 K) / ln(7.692L/4.442L)
Now simplify……
       W = 10087.6 J
       W = 10.08 kJ  

Hence work done of ideal gas by path A is 10.08 kJ

Learn more about ideal gas equation here : https://brainly.com/question/28837405?referrer=searchResults

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