Question

When 25.9 J was added as heat to a particular ideal gas, the volume of the gas changed from 41.0 cm3 to 67.5 cm3 while the pressure remained constant at 1.08 atm. (a) By how much did the internal energy of the gas change? If the quantity of gas present is 1.66 x 10-3 mol, find the molar specific heat of the gas at (b) constant pressure and (c) constant volume.

Answer 1

a)
`\Delta U=22.97J`

b)
`C_(p)=73.57(J)/(molK)`

c)
`C_(v)=65.25(J)/(molK)`

Step-by-step explanation:

`Q=25.9J`

`V_i=41.0cm^(3)=4.1e^(-5)m^3`

`V_f=67.5cm^(3)=6.75e^(-5)m^3`

`P=1.08atm=110444.3Pa`

a) The first law of thermodynamics tell us:

ΔU=Q-W

Where ΔU is the variation in internal energy, Q heat and W work done.

Now,
`W=P(V_f-V_i)`.

`\Delta U=Q-P(V_f-V_i)` ⇒
`\Delta U=22.97J`

b)
`n=1.66 \cdot  10^(-3) mol`

At constant pressure we can write:

`Q=nC_(p)\Delta T` ⇒
`C_(p)=(Q)/(n\Delta T)`

But what's the change in temperature? We can use the Ideal Gas Law:

`P\Delta V=nR\Delta T` ⇒
`\Delta T=(P\Delta V)/(nR) =212.05K`

∴
`C_(p)=73.57(J)/(molK)`

c) Now, we knoe that Cv and Cp are related by Cp-Cv=R.

⇒ Cv=Cp-R ⇒
`C_(v)=65.25(J)/(molK)`.