Question

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder is Pf=1.24 atm. What is the enthalpy change for this process? ΔH =

Answer 1

`\Delta H=-11897J`

Step-by-step explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

`\Delta H=\Delta U+V\Delta P`

Whereas the change in the internal energy is computed by:

`\Delta U=nCv\Delta T`

So we compute the initial and final temperatures for one mole of the ideal gas:

`T_1= (P_1V)/(nR)=(10.90atm*4.86L)/(0.082*n)=(646.02K )/(n) \\\\T_2= (P_2V)/(nR)=(1.24atm*4.86L)/(0.082*n)=(73.49K )/(n)`

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

`\Delta U=1mol*(3)/(2) (8.314(J)/(mol*K) )*(73.49K-646.02K )=-7140J`

Then, the volume-pressure product in Joules:

`V\Delta P=4.86L*(1m^3)/(1000L) *(1.24atm-10.90atm)*(101325Pa)/(1atm) \\\\V\Delta P=-4756.96J`

Finally, the change in the enthalpy for the process:

`\Delta H=-7140J-4757J\\\\\Delta H=-11897J`

Best regards.