Solution
1. System Analysis
The system consists of a piston-cylinder arrangement containing $CO_2$.
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 balances the weight of the piston. Since the piston mass ($M$) and area ($A$) are constant, the pressure $P = Mg/A$ is constant. This is an Isobaric Process.
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 ($PV = nRT$):
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 negligible volume of the solid)
3. Change in Internal Energy ($\Delta U$)
The total change in internal energy consists of two parts: the energy to sublimate the solid and the energy to heat the gases. $$ \Delta U = \Delta U_{\text{sublimation}} + \Delta U_{\text{heating}} $$
The total change in internal energy consists of two parts: the energy to sublimate the solid and the energy to heat the gases. $$ \Delta U = \Delta U_{\text{sublimation}} + \Delta U_{\text{heating}} $$
- Sublimation: The problem provides $L$ as the “molar specific heat of sublimation”. In the context of the First Law derivation here, this $L$ represents the internal energy required to break the bonds of the solid state ($n_d$ moles). $$ \Delta U_{\text{sublimation}} = n_d L $$
- Heating: The internal energy change of an ideal gas depends on $C_v$. We heat $(n_c + n_d)$ moles of gas from $T_c$ to $T$. $$ \Delta U_{\text{heating}} = (n_c + n_d) C_v (T – T_c) $$
4. Applying the First Law of Thermodynamics
$$ Q = \Delta U + W $$ Substitute the expressions from 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, let’s group the terms. Note that for an ideal gas $C_p = C_v + R$. Rearranging the Work term to match the temperature difference format: $$ W = (n_c + n_d)R(T – T_c) + n_d R T_c $$ *(Check: $(n_c+n_d)RT – (n_c+n_d)RT_c + n_d RT_c = (n_c+n_d)RT – n_c RT_c$. Correct.)* Now substituting this back into the heat equation: $$ Q = n_d L + (n_c + n_d)C_v(T – T_c) + (n_c + n_d)R(T – T_c) + n_d R T_c $$ $$ Q = n_d L + (n_c + n_d)(C_v + R)(T – T_c) + n_d R T_c $$ $$ Q = n_d L + (n_c + n_d)C_p(T – T_c) + n_d R T_c $$
$$ Q = \Delta U + W $$ Substitute the expressions from 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, let’s group the terms. Note that for an ideal gas $C_p = C_v + R$. Rearranging the Work term to match the temperature difference format: $$ W = (n_c + n_d)R(T – T_c) + n_d R T_c $$ *(Check: $(n_c+n_d)RT – (n_c+n_d)RT_c + n_d RT_c = (n_c+n_d)RT – n_c RT_c$. Correct.)* Now substituting this back into the heat equation: $$ Q = n_d L + (n_c + n_d)C_v(T – T_c) + (n_c + n_d)R(T – T_c) + n_d R T_c $$ $$ Q = n_d L + (n_c + n_d)(C_v + R)(T – T_c) + n_d R T_c $$ $$ Q = n_d L + (n_c + n_d)C_p(T – T_c) + n_d R T_c $$
5. Solving for Final Temperature ($T$)
Rearrange to isolate $T$: $$ (n_c + n_d) C_p (T – T_c) = Q – n_d L – n_d R T_c $$ Expand the LHS: $$ (n_c + n_d) C_p T – (n_c + n_d) C_p T_c = Q – n_d L – n_d R T_c $$ $$ (n_c + n_d) C_p T = Q – n_d L – n_d R T_c + n_c C_p T_c + n_d C_p T_c $$ $$ (n_c + n_d) C_p T = Q – n_d L + n_c C_p T_c + n_d (C_p – R) T_c $$ Since $C_p – R = C_v$: $$ (n_c + n_d) C_p T = Q – n_d L + n_c C_p T_c + n_d C_v T_c $$ Substitute values for $CO_2$:
$CO_2$ is a linear polyatomic molecule ($f=5$). $$ C_v = \frac{5}{2}R, \quad C_p = \frac{7}{2}R $$ Multiply the entire equation by 2 to clear fractions: $$ 2(n_c + n_d) \left(\frac{7}{2}R\right) T = 2Q – 2n_d L + 2n_c \left(\frac{7}{2}R\right) T_c + 2n_d \left(\frac{5}{2}R\right) T_c $$ $$ 7(n_c + n_d) R T = 2Q – 2n_d L + 7n_c R T_c + 5n_d R T_c $$ $$ 7(n_c + n_d) R T = 2Q – 2n_d L + (7n_c + 5n_d) R T_c $$ Finally, divide by the coefficient of $T$: $$ T = \frac{2Q – 2n_d L + (5n_d + 7n_c)RT_c}{7(n_c + n_d)R} $$
Rearrange to isolate $T$: $$ (n_c + n_d) C_p (T – T_c) = Q – n_d L – n_d R T_c $$ Expand the LHS: $$ (n_c + n_d) C_p T – (n_c + n_d) C_p T_c = Q – n_d L – n_d R T_c $$ $$ (n_c + n_d) C_p T = Q – n_d L – n_d R T_c + n_c C_p T_c + n_d C_p T_c $$ $$ (n_c + n_d) C_p T = Q – n_d L + n_c C_p T_c + n_d (C_p – R) T_c $$ Since $C_p – R = C_v$: $$ (n_c + n_d) C_p T = Q – n_d L + n_c C_p T_c + n_d C_v T_c $$ Substitute values for $CO_2$:
$CO_2$ is a linear polyatomic molecule ($f=5$). $$ C_v = \frac{5}{2}R, \quad C_p = \frac{7}{2}R $$ Multiply the entire equation by 2 to clear fractions: $$ 2(n_c + n_d) \left(\frac{7}{2}R\right) T = 2Q – 2n_d L + 2n_c \left(\frac{7}{2}R\right) T_c + 2n_d \left(\frac{5}{2}R\right) T_c $$ $$ 7(n_c + n_d) R T = 2Q – 2n_d L + 7n_c R T_c + 5n_d R T_c $$ $$ 7(n_c + n_d) R T = 2Q – 2n_d L + (7n_c + 5n_d) R T_c $$ Finally, divide by the coefficient of $T$: $$ T = \frac{2Q – 2n_d L + (5n_d + 7n_c)RT_c}{7(n_c + n_d)R} $$