Question

Three moles of an ideal gas (CV = 5 cal deg-1mol-1) at 10.0 atm and 00 are
converted to 2.0 atm at 500. Find ∆E and ∆H for the change.

Answer 1

1) $\Delta E=nCvdT$

n=3

T₂=50^oC=323,15K

T₁=0^oC=273.15K

$\Delta E=nCvdT=3\times5\times(323.15-273.15)=750 cal$

2)

$\Delta H=nCpdT=n(Cv+R)dT$

$\Delta H=n(Cv+R)dT=3\times(5+2)\times(323.15-273.15)=1050 cal$