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A gas is compressed from an initial volume of 5.40 L to a final volume of 1.23 L by an external pressure of 1.00 atm . During the compression the gas releases 128 J of heat.
My answer :
given data :
v1 = 5.40 l
v2 = 1.23 l
P1 - 1.00
now convert L to m^3 and atm to pa so
1 l =0.001 cubic meters
1 atm = 101,325 pa
v1 = 5.4X10^-3 m^3
v2 = 1.23 X 10^-3 m^3
p = 101325 pa
now DU = -Q - W
w = -pXDv = 101325(1.23 X 10^-3 - 5.4X10^-3 )
w = 422.5 J
now DU = -Q + W
= -128 + 422.5
Du = 294.5 J
My answer :
given data :
v1 = 5.40 l
v2 = 1.23 l
P1 - 1.00
now convert L to m^3 and atm to pa so
1 l =0.001 cubic meters
1 atm = 101,325 pa
v1 = 5.4X10^-3 m^3
v2 = 1.23 X 10^-3 m^3
p = 101325 pa
now DU = -Q - W
w = -pXDv = 101325(1.23 X 10^-3 - 5.4X10^-3 )
w = 422.5 J
now DU = -Q + W
= -128 + 422.5
Du = 294.5 J