THERMAL CYU 26

1. System Analysis

The system consists of a piston-cylinder arrangement containing $CO_2$.

• Initial State: $n_c$ moles of gas and $n_d$ moles of solid (dry ice) at temperature $T_c$.
• Final State: $(n_c + n_d)$ moles of gas at temperature $T$.
• Process: The external pressure is zero (vacuum). The pressure of the gas is solely responsible for balancing the weight of the piston ($Mg$). Since the piston mass and area are constant, the process is Isobaric.

2. Work Done ($W$)

Work is done by the gas as it expands against the constant pressure $P$.

$$ W = P(V_f – V_i) $$

Using the Ideal Gas Law ($PV = nRT$):

• Final Volume (all gas): $V_f = \frac{(n_c + n_d)RT}{P}$
• Initial Volume (gas portion): $V_i \approx \frac{n_c R T_c}{P}$ (Neglecting the volume of the solid)

Substituting these into the work equation:

$$ W = P \left[ \frac{(n_c + n_d)RT}{P} – \frac{n_c R T_c}{P} \right] $$ $$ W = (n_c + n_d)RT – n_c R T_c $$

3. Change in Internal Energy ($\Delta U$)

The total change in internal energy consists of two parts: the energy required to sublimate the solid dry ice and the energy required to raise the temperature of the gases.

$$ \Delta U = \Delta U_{\text{sublimation}} + \Delta U_{\text{heating}} $$

• Sublimation: $L$ is the molar specific internal energy of sublimation. $$ \Delta U_{\text{sublimation}} = n_d L $$
• Heating: We heat the total final amount of gas $(n_c + n_d)$ from the initial temperature $T_c$ to $T$. $$ \Delta U_{\text{heating}} = (n_c + n_d) C_v (T – T_c) $$

Combining these:

$$ \Delta U = n_d L + (n_c + n_d) C_v (T – T_c) $$

4. Applying the First Law of Thermodynamics

$$ Q = \Delta U + W $$

Substituting the expressions derived in steps 2 and 3:

$$ Q = \left[ n_d L + (n_c + n_d)C_v(T – T_c) \right] + \left[ (n_c + n_d)RT – n_c R T_c \right] $$

To simplify, we rearrange the Work term to match the temperature difference format. Note that $(n_c+n_d)RT – n_c RT_c$ can be rewritten as:

$$ W = (n_c + n_d)R(T – T_c) + n_d R T_c $$

Now substituting this back into the heat equation:

$$ Q = n_d L + (n_c + n_d)C_v(T – T_c) + (n_c + n_d)R(T – T_c) + n_d R T_c $$

Grouping the terms with $(T – T_c)$ and using Mayer’s Relation ($C_p = C_v + R$):

$$ Q = n_d L + (n_c + n_d)(C_v + R)(T – T_c) + n_d R T_c $$

5. Solving for Final Temperature ($T$)

Rearrange the equation to isolate $T$:

$$ (n_c + n_d) C_p (T – T_c) = Q – n_d L – n_d R T_c $$

Expanding the Left Hand Side:

$$ (n_c + n_d) C_p T – (n_c + n_d) C_p T_c = Q – n_d L – n_d R T_c $$ $$ (n_c + n_d) C_p T = Q – n_d L – n_d R T_c + n_c C_p T_c + n_d C_p T_c $$

Grouping the $T_c$ terms:

$$ (n_c + n_d) C_p T = Q – n_d L + n_c C_p T_c + n_d (C_p – R) T_c $$

Since $C_p – R = C_v$:

$$ (n_c + n_d) C_p T = Q – n_d L + n_c C_p T_c + n_d C_v T_c $$

Substituting values for $CO_2$:
$CO_2$ is a linear polyatomic molecule ($f=5$, ignoring vibrational modes at low temp).

$$ C_v = \frac{5}{2}R, \quad C_p = \frac{7}{2}R $$

Substituting these values:

$$ (n_c + n_d) \left(\frac{7}{2}R\right) T = Q – n_d L + n_c \left(\frac{7}{2}R\right) T_c + n_d \left(\frac{5}{2}R\right) T_c $$

Multiply by 2 to remove fractions:

$$ 7(n_c + n_d) R T = 2Q – 2n_d L + 7n_c R T_c + 5n_d R T_c $$ $$ 7(n_c + n_d) R T = 2Q – 2n_d L + (7n_c + 5n_d) R T_c $$

Final Expression: