- #1
duds1234
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Hello, this is my first post, so I apologize in advance if it's not quite up to par.
A gas is compressed at a constant pressure of 0.639 atm from 5.41 L to 2.3 L. In the process, 470 J of energy leaves the gas by heat.
a) What is the work done on the gas (J)?
b) What is the change in its internal energy(J)?
a) ##W=-PΔV##
b) ##ΔU=Q+W##
a)
##P = 0.639atm {\frac{1Pa}{9.872*10^-6atm}} = 64728.52512Pa##
##ΔV = V_f - V_i = 2.3L - 5.41L = -3.11L {\frac{1*10^{-3}m^3}{1L}} = -0.00311m^3##
##W = -P ΔV = -6.4728.52512Pa * -0.00311m^3 = 201.3057131J##
I submitted my answer to part a and it was correct. Part b is what I need help with. I believe that I am doing it correctly, but my answer is returned as incorrect. My incorrect work and answer appears below.
b)
##ΔU = Q + W = -407J + 201.3057131J = -205.6942869J##
Any insight into why my work/answer for part b is incorrect would be greatly appreciated!
Homework Statement
A gas is compressed at a constant pressure of 0.639 atm from 5.41 L to 2.3 L. In the process, 470 J of energy leaves the gas by heat.
a) What is the work done on the gas (J)?
b) What is the change in its internal energy(J)?
Homework Equations
a) ##W=-PΔV##
b) ##ΔU=Q+W##
The Attempt at a Solution
a)
##P = 0.639atm {\frac{1Pa}{9.872*10^-6atm}} = 64728.52512Pa##
##ΔV = V_f - V_i = 2.3L - 5.41L = -3.11L {\frac{1*10^{-3}m^3}{1L}} = -0.00311m^3##
##W = -P ΔV = -6.4728.52512Pa * -0.00311m^3 = 201.3057131J##
I submitted my answer to part a and it was correct. Part b is what I need help with. I believe that I am doing it correctly, but my answer is returned as incorrect. My incorrect work and answer appears below.
b)
##ΔU = Q + W = -407J + 201.3057131J = -205.6942869J##
Any insight into why my work/answer for part b is incorrect would be greatly appreciated!