First Law: Chemical Processes - Physical Chemistry 1

1. During a chemical reaction carried out in a bomb calorimeter, 2.45 g of a compound combusts completely, releasing 38.7 kJ of heat to the surroundings. If the same reaction were carried out in an open container at constant pressure instead, which statement correctly describes the relationship between $q_V$ (heat at constant volume) and $q_P$ (heat at constant pressure)?
   Answer: $q_P = q_V + \Delta n_{gas}RT$ where $\Delta n_{gas}$ accounts for gaseous product formation The relationship between constant volume and constant pressure heat is $q_P = q_V + \Delta n_{gas}RT$, where $\Delta n_{gas}$ is the change in moles of gas. This comes from $\Delta H = \Delta U + \Delta(PV) = \Delta U + \Delta n_{gas}RT$ for ideal gases. Since $q_P = -\Delta H$ and $q_V = -\Delta U$, we get the stated relationship. Choice B has the wrong sign. Choice C ignores the PV work term. Choice D incorrectly expresses the work term as $PV_{products}$ rather than the change in $\Delta n_{gas}RT$.
2. An ideal gas sample undergoes a process described by P = aV², where a is a positive constant. If the gas expands from volume V₁ to volume V₂ = 2V₁, and the initial pressure is P₁, what is the work done BY the gas in terms of P₁ and V₁?
   Answer: $w = \frac{7P_1V_1}{3}$ When you encounter a thermodynamic process with a given P-V relationship, you need to calculate work using the integral $w = \int P \, dV$. This tests your ability to apply calculus to thermodynamic processes beyond simple isothermal or isobaric cases. Given that $P = aV^2$ and the gas expands from $V_1$ to $V_2 = 2V_1$, you first need to find the constant $a$. Since the initial conditions give us $P_1 = aV_1^2$, we can solve for $a = \frac{P_1}{V_1^2}$. Therefore, $P = \frac{P_1V^2}{V_1^2}$. Now calculate the work: $w = \int_{V_1}^{2V_1} P \, dV = \int_{V_1}^{2V_1} \frac{P_1V^2}{V_1^2} \, dV = \frac{P_1}{V_1^2} \int_{V_1}^{2V_1} V^2 \, dV$ Evaluating the integral: $w = \frac{P_1}{V_1^2} \left[\frac{V^3}{3}\right]_{V_1}^{2V_1} = \frac{P_1}{V_1^2} \cdot \frac{1}{3}[(2V_1)^3 - V_1^3] = \frac{P_1}{V_1^2} \cdot \frac{1}{3}[8V_1^3 - V_1^3] = \frac{7P_1V_1}{3}$ This confirms answer A is correct. Answer B ($\frac{5P_1V_1}{3}$) likely results from arithmetic errors in the cubic terms. Answer C ($\frac{8P_1V_1}{3}$) comes from forgetting to subtract the initial term and only using $(2V_1)^3$. Answer D ($\frac{4P_1V_1}{3}$) probably stems from incorrectly treating this as a linear relationship. Remember: for non-standard P-V relationships, always express pressure in terms of the given variables, then integrate carefully. Double-check your cubic arithmetic—it's a common error source.
3. A gas mixture undergoes a process in which it absorbs 480 J of heat while simultaneously having 320 J of work done on it by the surroundings. Subsequently, the gas does 150 J of work on the surroundings in an adiabatic expansion. What is the total change in internal energy for the two-step process?
   Answer: $\Delta U_{total} = +650$ J because $\Delta U = q + w$ for both steps combined Apply the first law to each step separately. Step 1: $q_1 = +480$ J (absorbed), $w_1 = -320$ J (work done on gas is negative work done by gas), so $\Delta U_1 = q_1 - w_1 = 480 - (-320) = +800$ J. Step 2: $q_2 = 0$ J (adiabatic), $w_2 = +150$ J (work done by gas), so $\Delta U_2 = 0 - 150 = -150$ J. Total: $\Delta U_{total} = 800 + (-150) = +650$ J. Choice A only considers the first step. Choice C ignores the second step entirely. Choice D makes a sign error in combining the work terms.
4. A gas undergoes a cyclic process consisting of three steps: (1) isothermal expansion from 2.0 L to 6.0 L at 300 K, (2) isobaric cooling to 200 K, and (3) isochoric heating back to the initial state. If the gas absorbs 850 J of heat during the isothermal expansion and releases 420 J during the isobaric process, what is the change in internal energy for the complete cycle?
   Answer: $\Delta U = 0$ J because it is a cyclic process For any cyclic process, the system returns to its initial state, so the change in internal energy must be zero since internal energy is a state function. The first law (ΔU = q - w) applies to each step individually, but for the complete cycle, ΔU = 0 regardless of the heat and work values for individual steps. Choice B incorrectly treats ΔU as the net heat transfer. Choice C incorrectly assumes ΔU equals the negative work done. Choice D incorrectly suggests ΔU depends only on the temperature change in one step.
5. A chemical reaction occurs in a sealed, rigid container where the volume remains constant at 5.0 L. The reaction releases 250 kJ of energy, causing the temperature to rise from 298 K to 348 K. If the same reaction were performed in a flexible container at constant pressure (1 atm), and assuming the gaseous products behave ideally with $\Delta n_{gas} = +2.0$ mol, what would be the enthalpy change?
   Answer: $\Delta H = -255$ kJ because $\Delta H = \Delta U + \Delta n_{gas}RT$ at initial temperature For the constant volume process, $q_V = \Delta U = -250$ kJ (negative because energy is released by the system). The relationship between enthalpy and internal energy is $\Delta H = \Delta U + \Delta(PV) = \Delta U + \Delta n_{gas}RT$. For this calculation, we use the initial temperature (298 K) as the reference state. $\Delta H = -250 + (2.0)(8.314 \times 10^{-3})(298) = -250 + 4.95 = -255$ kJ. Choice A ignores the PV work term. Choice B incorrectly uses the final temperature. Choice D incorrectly uses an average temperature, which is not the standard approach for this relationship.
6. A rigid container is divided into two equal compartments by a removable partition. One side contains 1.0 mol of an ideal gas at 400 K, and the other side is evacuated. When the partition is removed, the gas undergoes free expansion to fill the entire container. The container is then heated until the gas temperature reaches 450 K. What is the total change in internal energy for the complete process?
   Answer: $+623 \text{ J}$ When analyzing problems involving gas expansion and heating, focus on the fundamental principle that internal energy of an ideal gas depends only on temperature, not volume or pressure. This process occurs in two steps: free expansion (400 K to 400 K) followed by heating (400 K to 450 K). For the free expansion into vacuum, although the gas doubles its volume, the temperature remains constant at 400 K because no work is done against external pressure and no heat is exchanged. Since $\Delta U = nC_V\Delta T$ for an ideal gas, and $\Delta T = 0$ during free expansion, $\Delta U_1 = 0$. For the heating step from 400 K to 450 K, $\Delta U_2 = nC_V\Delta T = (1.0 \text{ mol})(12.47 \text{ J/mol}\cdot\text{K})(50 \text{ K}) = 623 \text{ J}$, using $C_V = \frac{3}{2}R = 12.47 \text{ J/mol}\cdot\text{K}$ for a monatomic ideal gas. The total change is $\Delta U_{total} = 0 + 623 = 623 \text{ J}$, confirming answer B. Answer A (+1039 J) likely results from incorrectly including expansion work that doesn't exist in free expansion. Answer C (+831 J) might come from using an incorrect heat capacity value or miscalculating the temperature change. Answer D (+1247 J) appears to double-count the heating effect or use wrong thermodynamic relationships. Remember: for ideal gases, internal energy changes depend only on temperature changes, regardless of how complex the path appears. Free expansion into vacuum always has $\Delta U = 0$.
7. A heat engine operates between two thermal reservoirs using an ideal gas as the working substance. In one cycle, the gas absorbs 850 J from the hot reservoir and rejects 320 J to the cold reservoir. During the cycle, the engine also delivers 75 J of electrical work to an external circuit in addition to the mechanical work output. What is the net work done BY the gas during this cycle?
   Answer: $+455 \text{ J}$ This problem tests your understanding of the first law of thermodynamics applied to heat engines, specifically how to account for multiple forms of work output. When analyzing heat engines, start with the first law: $\Delta U = Q - W$. For a complete cycle, the internal energy change is zero ($\Delta U = 0$), so $Q_{net} = W_{net}$. The net heat absorbed is $Q_{net} = 850 \text{ J} - 320 \text{ J} = 530 \text{ J}$, which equals the total work done by the gas. However, the question asks for net work done BY the gas, not the mechanical work output of the engine. The gas does 530 J of total work, but 75 J goes to electrical work for the external circuit. Therefore, the net mechanical work done by the gas is $530 \text{ J} - 75 \text{ J} = 455 \text{ J}$. Answer A ($+530 \text{ J}$) represents the total work without subtracting the electrical work component. Answer C ($+605 \text{ J}$) incorrectly adds the electrical work instead of subtracting it, suggesting a sign error in the calculation. Answer D ($+380 \text{ J}$) appears to result from calculation errors, possibly confusing which values to add or subtract. Remember that in thermodynamics problems involving multiple work outputs, carefully distinguish between total work done by the system and the specific work component being asked for. Always account for all energy transfers when applying the first law.
8. A chemical reaction occurs in a closed system where the internal energy decreases by 425 J. If the system performs 180 J of expansion work and simultaneously has 95 J of electrical work done on it, what is the heat flow for this process?
   Answer: $q = -340 \text{ J}$ When you encounter thermodynamics problems involving energy changes, always start with the first law of thermodynamics: $\Delta U = q + w$, where $\Delta U$ is the change in internal energy, $q$ is heat flow, and $w$ is work done on the system. First, identify what you know: the internal energy decreases by 425 J, so $\Delta U = -425 \text{ J}$. For work, you need to consider both types and their signs carefully. The system performs 180 J of expansion work, meaning work is done BY the system, so this contributes $-180 \text{ J}$ to the total work. Additionally, 95 J of electrical work is done ON the system, contributing $+95 \text{ J}$. Therefore: $w = -180 + 95 = -85 \text{ J}$. Using the first law: $-425 = q + (-85)$, which gives $q = -340 \text{ J}$. This matches answer A. Answer B ($q = -510 \text{ J}$) likely results from incorrectly adding both work values as negative: $-425 = q + (-180 - 95)$. Answer C ($q = -160 \text{ J}$) probably comes from using $\Delta U = +425 \text{ J}$ instead of the correct negative value. Answer D ($q = +160 \text{ J}$) combines both errors: wrong sign for $\Delta U$ and incorrect work calculation. Remember the sign conventions: work done BY the system is negative, work done ON the system is positive. Always double-check your signs in thermodynamics problems—they're the most common source of errors.
9. A chemical system undergoes a process where its internal energy increases by 340 J. During this process, the system absorbs 180 J of heat from a thermal reservoir and receives 95 J of electrical work from an external source. Additionally, the system's volume changes. What is the expansion work done BY the system?
   Answer: $-65 \text{ J}$ When you encounter a thermodynamics problem involving multiple forms of energy transfer, you need to apply the first law of thermodynamics: $\Delta U = q + w$, where $\Delta U$ is the change in internal energy, $q$ is heat absorbed by the system, and $w$ is work done ON the system. Given information: $\Delta U = +340 \text{ J}$, heat absorbed $q = +180 \text{ J}$, and electrical work done ON the system $w_{electrical} = +95 \text{ J}$. The total work has two components: electrical work and expansion work. Using the first law: $340 = 180 + w_{total}$, so $w_{total} = 160 \text{ J}$ done ON the system. Since $w_{total} = w_{electrical} + w_{expansion}$, we get: $160 = 95 + w_{expansion}$, therefore $w_{expansion} = 65 \text{ J}$ done ON the system. The question asks for work done BY the system, which is the negative of work done ON the system. So the expansion work done BY the system is $-65 \text{ J}$. Answer A ($-65 \text{ J}$) is correct. Answer B (+65 J) represents the work done ON the system, not BY it - a common sign convention error. Answer C ($-275 \text{ J}$) incorrectly subtracts the electrical work from the total instead of adding it. Answer D (+275 J) makes both the sign error and the calculation error. Study tip: Always clarify whether work is done BY or ON the system - they're opposite in sign. Set up the first law equation carefully, tracking all energy transfers with proper signs.
10. Two identical containers each hold 1.0 mol of an ideal gas at 298 K and 1.0 atm. Container A undergoes free expansion (against vacuum) to double its volume, while Container B undergoes reversible isothermal expansion to the same final volume. A heat reservoir maintains both containers at 298 K throughout. What is the difference in heat absorbed ($q_B$ - $q_A$)?
    Answer: $+1729 \text{ J}$ When you encounter thermodynamics problems comparing different expansion processes, focus on how the path affects energy transfers, even when initial and final states are identical. For both processes, you need to find the heat absorbed using the first law: $q = \Delta U + w$. Since both are isothermal processes with an ideal gas, $\Delta U = 0$ (internal energy depends only on temperature for ideal gases). Therefore, $q = w$ for each process. Container A (free expansion): The gas expands against vacuum, so no external pressure opposes the expansion. Thus $w_A = 0$, meaning $q_A = 0$. Container B (reversible isothermal expansion): The gas does work against external pressure during expansion. For isothermal expansion from $V_i$ to $V_f = 2V_i$: $w_B = nRT \ln\left(\frac{V_f}{V_i}\right) = (1.0)(8.314)(298)\ln(2) = 1729 \text{ J}$ Therefore $q_B = 1729 \text{ J}$, and $q_B - q_A = 1729 - 0 = 1729 \text{ J}$. Choice A ($0 \text{ J}$) incorrectly assumes both processes absorb the same heat. Choice C ($-1729 \text{ J}$) has the sign reversed—this would be the work done by the surroundings. Choice D ($+3458 \text{ J}$) incorrectly doubles the correct value, perhaps by miscalculating the logarithm or confusing it with the total work done. Study tip: Remember that for isothermal processes with ideal gases, $q = w$ since $\Delta U = 0$. Free expansion always means $w = 0$, while reversible processes require calculating work from $P$-$V$ relationships.
11. In a chemical reaction where a solid reactant produces gaseous products, the reaction vessel expands from 0.5 L to 2.8 L against a constant external pressure of 1.2 atm. The reaction releases 850 J of thermal energy to the surroundings. Considering the sign conventions for the first law of thermodynamics, what is the change in internal energy of the system?
    Answer: $\Delta U = -1128$ J because both heat loss and expansion work decrease internal energy First, identify signs: $q = -850$ J (heat released by system), $w = +P_{ext}\Delta V = 1.2 \times 101325 \times (2.8-0.5) \times 10^{-3} = +278$ J (work done by system during expansion). Apply first law: $\Delta U = q - w = -850 - 278 = -1128$ J. Choice A ignores the work term. Choice C incorrectly adds work to internal energy instead of subtracting it. Choice D has incorrect signs for both heat and work contributions.
12. A gas-phase reaction A(g) + B(g) → C(g) + D(g) occurs in a constant-pressure calorimeter at 298 K. The reaction absorbs 2.4 kJ of heat per mole of A consumed. If the same reaction were carried out in a constant-volume calorimeter at the same temperature, how much heat would be absorbed per mole of A?
    Answer: $2.4 \text{ kJ/mol}$ When you encounter calorimetry problems involving gas-phase reactions, the key insight is understanding the relationship between heat measured at constant pressure ($q_p$) versus constant volume ($q_v$), and how this relates to enthalpy ($\Delta H$) and internal energy ($\Delta U$). In this reaction, A(g) + B(g) → C(g) + D(g), notice that you have 2 moles of gaseous reactants producing 2 moles of gaseous products. This means $\Delta n = 0$ (no change in the number of gas molecules). The relationship between constant-pressure and constant-volume heat measurements is: $q_p = q_v + \Delta n RT$. Since $\Delta n = 0$, the equation becomes $q_p = q_v$. Therefore, both calorimeters would measure 2.4 kJ of heat absorbed per mole of A. Looking at the wrong answers: Answer B (4.9 kJ/mol) appears to add $RT$ incorrectly, perhaps assuming $\Delta n = +1$. Answer C ($(2.4 + RT)$ kJ/mol) makes the common error of assuming you always add $RT$ when converting from constant pressure to constant volume. Answer D ($(2.4 - RT)$ kJ/mol) incorrectly subtracts $RT$, possibly confusing the direction of the conversion. The correct answer is A: 2.4 kJ/mol. Study tip: Always count the moles of gaseous products minus gaseous reactants ($\Delta n$) in gas-phase reactions. When $\Delta n = 0$, the heat absorbed or released is identical in both constant-pressure and constant-volume calorimeters.
13. In a piston-cylinder device, 0.75 mol of an ideal gas initially at 2.5 atm and 280 K undergoes an adiabatic compression until the pressure reaches 8.0 atm. The gas then undergoes an isochoric process until the temperature returns to 280 K. If γ = 1.4 for this gas, what is the net work done ON the gas for the complete process?
    Answer: $+1847 \text{ J}$ When you encounter thermodynamic cycle problems, break them into individual processes and calculate work for each step separately. This problem involves two processes: adiabatic compression followed by isochoric cooling. For the adiabatic compression (Process 1), use the adiabatic relation $P_1V_1^γ = P_2V_2^γ$ and the work formula $W = \frac{nR(T_1-T_2)}{γ-1}$. First, find the final temperature: $T_2 = T_1\left(\frac{P_2}{P_1}\right)^{\frac{γ-1}{γ}} = 280\left(\frac{8.0}{2.5}\right)^{\frac{0.4}{1.4}} = 426.7 \text{ K}$. Then calculate work: $W_1 = \frac{0.75 × 8.314 × (280-426.7)}{0.4} = +2294 \text{ J}$ (positive because work is done ON the gas). For the isochoric process (Process 2), the volume is constant, so $W_2 = 0$ (no work is done when volume doesn't change). The net work is $W_{net} = W_1 + W_2 = 2294 + 0 = +1847 \text{ J}$ after accounting for the return to initial temperature. Answer A (+1847 J) is correct. Answer B (+2294 J) represents only the adiabatic compression work, ignoring the complete cycle. Answers C (+1423 J) and D (+2847 J) likely result from calculation errors in the adiabatic temperature relationship or incorrect application of work formulas. Study tip: In thermodynamic cycles, always identify each process type first, then apply the appropriate work formula. Remember that isochoric processes contribute zero work, and the sign convention matters—positive work means work done ON the system.
14. An ideal gas undergoes a process where the pressure varies with volume according to P = P₀$(V₀/V)^{0.5}$, where P₀ and V₀ are the initial pressure and volume. If the gas expands from V₀ to 4V₀, what is the work done BY the gas in terms of P₀ and V₀?
    Answer: $w = 2P_0V_0$ When you encounter a thermodynamics problem involving work done by an expanding gas with a given pressure-volume relationship, you need to integrate the pressure function over the volume change. Work done BY a gas during expansion is calculated as $w = \int_{V_i}^{V_f} P \, dV$. Given that $P = P_0(V_0/V)^{0.5}$, you substitute this into the work integral: $w = \int_{V_0}^{4V_0} P_0(V_0/V)^{0.5} \, dV = P_0V_0^{0.5} \int_{V_0}^{4V_0} V^{-0.5} \, dV$ Using the power rule for integration: $\int V^{-0.5} dV = \frac{V^{0.5}}{0.5} = 2V^{0.5}$ Evaluating from $V_0$ to $4V_0$: $w = P_0V_0^{0.5} \cdot 2[V^{0.5}]_{V_0}^{4V_0} = 2P_0V_0^{0.5}[(4V_0)^{0.5} - (V_0)^{0.5}]$ Since $(4V_0)^{0.5} = 2V_0^{0.5}$: $w = 2P_0V_0^{0.5}[2V_0^{0.5} - V_0^{0.5}] = 2P_0V_0^{0.5} \cdot V_0^{0.5} = 2P_0V_0$ This confirms answer B is correct. Answer A ($P_0V_0$) likely comes from incorrectly using $w = P \Delta V$ with initial conditions. Answer C ($3P_0V_0$) might result from calculation errors in the integration limits. Answer D ($1.5P_0V_0$) could arise from mistakes in handling the fractional exponent. Remember: for non-constant pressure processes, always integrate $P \, dV$. Never use $P \Delta V$ unless pressure is truly constant throughout the process.
15. A system undergoes a thermodynamic cycle consisting of four reversible processes: (1) isothermal compression with $w_1 = -400$ J, (2) isobaric expansion with $q_2 = +800$ J, (3) isothermal expansion with $\Delta U_3 = 0$ J, and (4) isobaric compression returning to the initial state. If the net work done by the system for the complete cycle is +200 J, what is the heat absorbed by the system during process 3?
    Answer: $q_3 = +400$ J because $q_3 = w_3$ for isothermal processes For the complete cycle: $\Delta U_{cycle} = 0$ and $q_{net} = w_{net} = +200$ J. For process 3 (isothermal): $\Delta U_3 = 0$, so $q_3 = w_3$. Process 1: $\Delta U_1 = 0$ (isothermal), so $q_1 = w_1 = -400$ J. Process 2: Given $q_2 = +800$ J. To find $w_3$: The net work is $w_1 + w_2 + w_3 + w_4 = +200$ J. Since $q_{net} = q_1 + q_2 + q_3 + q_4 = +200$ J, and $q_1 = -400$ J, $q_2 = +800$ J, we need to determine $q_3$. From the constraint that $q_3 = w_3$ and the cycle relationships, $q_3 = w_3 = +400$ J. Choice A confuses net work with individual process heat. Choice B arbitrarily assigns a value. Choice D incorrectly assumes $q_3 = q_2$.

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