Question

A mixture of nitrogen and hydrogen gas is expanded from a volume of 55.0 L to a volume of 66.0 L, while the pressure is held constant at 67.0 atm. Calc the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits. x 5 ?

Answer 1

Answer : 74.67 kJ, 7.46 *10^1 kJ

explanation :

The initial volume of the mixture gas V1 = 55.0 L

The final volume of the mixture of gas V2= 66.0 L

The expansion in Volume of mixture of Gases ∆V= V2-V1

=( 66.0 L - 55.0 L) =11.0 L

Pressure P =67.0 atm,

Thus the work done on gas mixture at constant pressure W = -P.∆V

W = -(67.0 atm * 11 L)

W= - 737 L.atm

Now to convert L.atm into joule ,

W = -737 *101.25 J /1 L.atm

W = 74676.52 J

W = 74.676 KJ

W = 7.46 * 10^1 KJ

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