### Finding Q,W,U,H when Heat Capacity of a Gas Is a Function of T, Expansion (First Law of Thermodynamics) (Example 3)

Sometimes you will find thermodynamics questions with gas expansion, but the heat capacity of the gas is not a constant; instead, it is a function of temperature. These questions aren't too difficult, you just need to integrate the function of Cp or Cv with respect to T in your thermodynamics equations.

The molar heat capacity of a gas is approximated as:

${C_{p,m}} = 6.15\frac{{cal}}{{molK}} + \left( {0.0031\frac{{cal}}{{mol{K^2}}}} \right)(T)$

Assuming ideal gas behaviour:

a.) Find Q, W, $$\Delta H$$, $$\Delta U$$ when 2 mol of this gas is reversibly heated from 27 C to 127 C, with P fixed at 1 atm. (isobaric)

b.) Find Q, W, $$\Delta H$$, $$\Delta U$$ when 2 mol of this gas initially at 1 atm is reversibly heated from 27 C to 127 C with volume being fixed. (isochloric)

Remember the relation between Cvm and Cpm, which holds true even if C is a function of T. Also use the Kelvin temperatures as the integral bounds.

${C_{v,m}} = {C_{p,m}} - R$

$\Delta U = n\int {{C_{v,m}}} dT$

$\Delta U = 2\int_{300}^{400} {(0.0031T + 4.1641)dT}$

$\Delta U = 1050cal$

$\Delta H = n\int {{C_{p,m}}} dT$

$\Delta H = 2\int_{300}^{400} {(0.0031T + 6.15)dT}$

$\Delta H = 1447cal$

$W = - nR\Delta T$

$W = - (2mol)\left( {8.314\frac{J}{{molK}}} \right)(400K - 300K)$

$W = - 1662.80J = - 397cal$

For isobaric expansion:

$Q = \Delta U - W = \Delta H$

Q = 1447 cal

Now let's do part b, which is constant volume or isochloric expansion.

Since V is fixed, W = 0.

$\Delta U = n\int {{C_{v,m}}} dT$

$\Delta U = 2\int_{300}^{400} {(0.0031T + 4.1641)dT}$

$\Delta U = 1050cal$

$\Delta H = n\int {{C_{p,m}}} dT$

$\Delta H = 2\int_{300}^{400} {(0.0031T + 6.15)dT}$

$\Delta H = 1447cal$

And since the V is fixed (isochloric):

$Q = \Delta U = 1050cal$

Now we solved for all the variables in part b.)!