Question 1 of 7: First-Law Bookkeeping for an Ideal Gas — Irreversible Expansion Then Isochoric Cooling
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 10-Met-A1 Metallurgical Thermodynamics. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator; candidates were told to state any interpretive assumptions. Any five of the seven questions constitute a complete paper — all seven are solved below for completeness. All questions are of equal value (20 marks each out of 100).
Reference texts: Gaskell, D. R., Introduction to the Thermodynamics of Materials (2nd–5th ed.) — the exam's own Ellingham diagram (Fig. 9-3) is reproduced from this text, which also supplies the standard-state enthalpy/entropy data used below; supporting 298 K entropies for the elements, CO, H₂, H₂O(g) and CO₂ from the NIST-JANAF Thermochemical Tables.
Question 1: First-Law Bookkeeping for an Ideal Gas — Irreversible Expansion Then Isochoric Cooling (20 marks)
Given. $n=1$ mol of an ideal gas; state 1 at $T_1=300$ K, $P_1=10^5$ Pa. (a) Isothermal, irreversible expansion against a constant external pressure $P_{ext}=10^4$ Pa until the volume doubles (state 2). (b) Constant-volume cooling from 300 K to 250 K (state 3).
Check
The problem does not name the gas, so it is taken as monatomic ($C_v=\tfrac32R$, $C_p=\tfrac52R$) — the simplest "ideal gas" case. For a diatomic gas ($C_v=\tfrac52R$, $C_p=\tfrac72R$) the constant-volume $\Delta U$, $\Delta H$ and $q$ in part (b)/(c) below would scale by $5/3$; part (a) is unaffected, since it depends only on $P_{ext}\Delta V$.
Find. $q$, $w$, $\Delta U$, $\Delta H$ for step (a), step (b), and the overall two-step process (c).
Figure 1 — State path on a P–V diagram: (a) irreversible isothermal expansion at constant $P_{ext}=10\ \text{kPa}$ from state 1 to state 2 (dotted curve is the reversible 300 K isotherm for reference, not the actual path); (b) isochoric cooling from state 2 to state 3.
Approach. Fix state 2 from the ideal-gas law at constant $T$, get the irreversible work from the constant opposing pressure ($w=-P_{ext}\Delta V$), get $\Delta U$ and $\Delta H$ from the temperature change only (both are functions of $T$ alone for an ideal gas), and close each step with the first law $q=\Delta U-w$.
State 1 and state 2 volumes. $V_1=\dfrac{nRT_1}{P_1}=\dfrac{(1)(8.314)(300)}{10^5}=0.024942\ \text{m}^3=24.94\ \text{L}$, so $V_2=2V_1=49.88\ \text{L}$. Since (a) is isothermal, the ideal-gas law fixes the state-2 pressure at $T_1=300$ K: $P_2=nRT_1/V_2=P_1/2=50\ \text{kPa}$ — note this is not equal to $P_{ext}=10\ \text{kPa}$, because the process is irreversible and only the initial and final states need lie on the 300 K isotherm.
Part (a) — work and heat. The gas pushes back a constant external pressure, so $w=-P_{ext}(V_2-V_1) = -(10^4)(0.024942) = \boxed{-249.4\ \text{J}}$ (work done on the system, IUPAC sign convention). Because $T$ is constant for an ideal gas, $\Delta U_a=0$ and $\Delta H_a=0$; the first law then gives $q_a=\Delta U_a-w_a=\boxed{+249.4\ \text{J}}$.
Part (b) — constant-volume cooling. No boundary moves, so $w_b=0$. With $\Delta T=250-300=-50$ K, $\Delta U_b=nC_v\Delta T=(1)(1.5)(8.314)(-50)=\boxed{-623.6\ \text{J}}=q_b$ (since $w_b=0$), and $\Delta H_b=nC_p\Delta T=(1)(2.5)(8.314)(-50)=\boxed{-1039.3\ \text{J}}$. (State 3 pressure: $P_3=P_2\,T_3/T_2=50\times250/300=41.7$ kPa, at the same $V_2$.)
Part (c) — the combined process. All four quantities are additive over the two steps: $q_c=q_a+q_b=249.4-623.6=\boxed{-374.1\ \text{J}}$; $w_c=w_a+w_b=-249.4\ \text{J}$; $\Delta U_c=\Delta U_a+\Delta U_b=-623.6\ \text{J}$; $\Delta H_c=\Delta H_a+\Delta H_b=-1039.3\ \text{J}$.
