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physics help please

A gas is compressed at a constant pressure of 0.800 atm from 9.00 Lto 1.00 L. In the process,420 J of energy leaves the gas byheat.(a) What is the work done on thegas?
J
(b) What is the change in its internal energy?
J
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Answer #1
GivenPressure P = 0.800 atm= 0.8*( 1.013*105)Painitial volume Vi=9.00 L=9*10-3m3final volume Vf = 1.00 L =1*10-3m3heat dQ = 420 J--------------------------------------------------------------------------------------a) the work doneon the gas is given bydW = - P dVdW= - P (Vf -Vi)Plug in values-------------------------------------------------------------------b)thechange in its internal energy is given byFrom the first law of thermodynamicsdQ = dU + dWdU = dQ - dWdU= 420 - [ - P ( Vf-Vi)]Plug in values
answered by: kennisha
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