Three moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L h

Question

Three moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 90.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 178 J.

WIAF = J WIBF = J WIF = J Three moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 90.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 178 J. For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas. For the paths IAF, IBF, and IF in the figure above, calculate the net energy transferred to the gas by heat in the process.

Explanation / Answer

work done is area under the curve so for path IAF W=1.5*0.5=0.75 l atm=76 J
for path IF W=1.5*0.5+0.5*0.5*0.5=0.875 l atm=88.66 J
for path IBF W=2*0.5=1 l atm=101.33 J

q=w+(u2-u1)
u2-u1=178-90=88J
for IAF q=76+88=164 J
for IF q=88.66+88=176,6 J
for IBV q=101.33+88=189.3 J

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