Physical Chemistry 1 Quiz: First Law Chemical Processes
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First Law Chemical ProcessesQuestion 1 of 15

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?

+1847 J+1847 \text{ J}
+2294 J+2294 \text{ J}
+1423 J+1423 \text{ J}
+2847 J+2847 \text{ J}
+3294 J+3294 \text{ J}
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Physical Chemistry 1 Quiz

Physical Chemistry 1 Quiz: First Law Chemical Processes

Practice First Law Chemical Processes in Physical Chemistry 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on First Law Chemical Processes, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

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?

  1. +1847 J+1847 \text{ J} (correct answer)
  2. +2294 J+2294 \text{ J}
  3. +1423 J+1423 \text{ J}
  4. +2847 J+2847 \text{ J}
  5. +3294 J+3294 \text{ J}
Explanation: 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 P1V1γ=P2V2γP_1V_1^γ = P_2V_2^γ and the work formula W=nR(T1−T2)γ−1W = \frac{nR(T_1-T_2)}{γ-1}. First, find the final temperature: T2=T1(P2P1)γ−1γ=280(8.02.5)0.41.4=426.7 KT_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: W1=0.75×8.314×(280−426.7)0.4=+2294 JW_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 W2=0W_2 = 0 (no work is done when volume doesn't change). The net work is Wnet=W1+W2=2294+0=+1847 JW_{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.

Question 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?

  1. w=7P1V13w = \frac{7P_1V_1}{3} (correct answer)
  2. w=5P1V13w = \frac{5P_1V_1}{3}
  3. w=8P1V13w = \frac{8P_1V_1}{3}
  4. w=4P1V13w = \frac{4P_1V_1}{3}
  5. w=2P1V13w = \frac{2P_1V_1}{3}
Explanation: When you encounter a thermodynamic process with a given P-V relationship, you need to calculate work using the integral w=∫P dVw = \int P \, dV. This tests your ability to apply calculus to thermodynamic processes beyond simple isothermal or isobaric cases. Given that P=aV2P = aV^2 and the gas expands from V1V_1 to V2=2V1V_2 = 2V_1, you first need to find the constant aa. Since the initial conditions give us P1=aV12P_1 = aV_1^2, we can solve for a=P1V12a = \frac{P_1}{V_1^2}. Therefore, P=P1V2V12P = \frac{P_1V^2}{V_1^2}. Now calculate the work: w=∫V12V1P dV=∫V12V1P1V2V12 dV=P1V12∫V12V1V2 dVw = \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=P1V12[V33]V12V1=P1V12⋅13[(2V1)3−V13]=P1V12⋅13[8V13−V13]=7P1V13w = \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 (5P1V13\frac{5P_1V_1}{3}) likely results from arithmetic errors in the cubic terms. Answer C (8P1V13\frac{8P_1V_1}{3}) comes from forgetting to subtract the initial term and only using (2V1)3(2V_1)^3. Answer D (4P1V13\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.

Question 3

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?

  1. 2.4 kJ/mol2.4 \text{ kJ/mol} (correct answer)
  2. 4.9 kJ/mol4.9 \text{ kJ/mol}
  3. (2.4+RT) kJ/mol(2.4 + RT) \text{ kJ/mol}
  4. (2.4−RT) kJ/mol(2.4 - RT) \text{ kJ/mol}
  5. 0.0 kJ/mol0.0 \text{ kJ/mol}
Explanation: When you encounter calorimetry problems involving gas-phase reactions, the key insight is understanding the relationship between heat measured at constant pressure (qpq_p) versus constant volume (qvq_v), and how this relates to enthalpy (ΔH\Delta H) and internal energy (ΔU\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 Δn=0\Delta n = 0 (no change in the number of gas molecules). The relationship between constant-pressure and constant-volume heat measurements is: qp=qv+ΔnRTq_p = q_v + \Delta n RT. Since Δn=0\Delta n = 0, the equation becomes qp=qvq_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 RTRT incorrectly, perhaps assuming Δn=+1\Delta n = +1. Answer C ((2.4+RT)(2.4 + RT) kJ/mol) makes the common error of assuming you always add RTRT when converting from constant pressure to constant volume. Answer D ((2.4−RT)(2.4 - RT) kJ/mol) incorrectly subtracts RTRT, 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 (Δn\Delta n) in gas-phase reactions. When Δn=0\Delta n = 0, the heat absorbed or released is identical in both constant-pressure and constant-volume calorimeters.

Question 4

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?

  1. +1039 J+1039 \text{ J}
  2. +623 J+623 \text{ J} (correct answer)
  3. +831 J+831 \text{ J}
  4. +1247 J+1247 \text{ J}
  5. +519 J+519 \text{ J}
Explanation: 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 ΔU=nCVΔT\Delta U = nC_V\Delta T for an ideal gas, and ΔT=0\Delta T = 0 during free expansion, ΔU1=0\Delta U_1 = 0. For the heating step from 400 K to 450 K, ΔU2=nCVΔT=(1.0 mol)(12.47 J/mol\cdotpK)(50 K)=623 J\Delta U_2 = nC_V\Delta T = (1.0 \text{ mol})(12.47 \text{ J/mol·K})(50 \text{ K}) = 623 \text{ J}, using CV=32R=12.47 J/mol\cdotpKC_V = \frac{3}{2}R = 12.47 \text{ J/mol·K} for a monatomic ideal gas. The total change is ΔUtotal=0+623=623 J\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 ΔU=0\Delta U = 0.

Question 5

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?

  1. w=P0V0w = P_0V_0
  2. w=2P0V0w = 2P_0V_0 (correct answer)
  3. w=3P0V0w = 3P_0V_0
  4. w=1.5P0V0w = 1.5P_0V_0
  5. w=0.5P0V0w = 0.5P_0V_0
Explanation: 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=∫ViVfP dVw = \int_{V_i}^{V_f} P \, dV. Given that P=P0(V0/V)0.5P = P_0(V_0/V)^{0.5}, you substitute this into the work integral: w=∫V04V0P0(V0/V)0.5 dV=P0V00.5∫V04V0V−0.5 dVw = \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: ∫V−0.5dV=V0.50.5=2V0.5\int V^{-0.5} dV = \frac{V^{0.5}}{0.5} = 2V^{0.5} Evaluating from V0V_0 to 4V04V_0: w=P0V00.5⋅2[V0.5]V04V0=2P0V00.5[(4V0)0.5−(V0)0.5]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 (4V0)0.5=2V00.5(4V_0)^{0.5} = 2V_0^{0.5}: w=2P0V00.5[2V00.5−V00.5]=2P0V00.5⋅V00.5=2P0V0w = 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 (P0V0P_0V_0) likely comes from incorrectly using w=PΔVw = P \Delta V with initial conditions. Answer C (3P0V03P_0V_0) might result from calculation errors in the integration limits. Answer D (1.5P0V01.5P_0V_0) could arise from mistakes in handling the fractional exponent. Remember: for non-constant pressure processes, always integrate P dVP \, dV. Never use PΔVP \Delta V unless pressure is truly constant throughout the process.

Question 6

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?

  1. −65 J-65 \text{ J} (correct answer)
  2. +65 J+65 \text{ J}
  3. −275 J-275 \text{ J}
  4. +275 J+275 \text{ J}
  5. −160 J-160 \text{ J}
Explanation: When you encounter a thermodynamics problem involving multiple forms of energy transfer, you need to apply the first law of thermodynamics: ΔU=q+w\Delta U = q + w, where ΔU\Delta U is the change in internal energy, qq is heat absorbed by the system, and ww is work done ON the system. Given information: ΔU=+340 J\Delta U = +340 \text{ J}, heat absorbed q=+180 Jq = +180 \text{ J}, and electrical work done ON the system welectrical=+95 Jw_{electrical} = +95 \text{ J}. The total work has two components: electrical work and expansion work. Using the first law: 340=180+wtotal340 = 180 + w_{total}, so wtotal=160 Jw_{total} = 160 \text{ J} done ON the system. Since wtotal=welectrical+wexpansionw_{total} = w_{electrical} + w_{expansion}, we get: 160=95+wexpansion160 = 95 + w_{expansion}, therefore wexpansion=65 Jw_{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 J-65 \text{ J}. Answer A (−65 J-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 J-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.

Question 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?

  1. +530 J+530 \text{ J}
  2. +455 J+455 \text{ J} (correct answer)
  3. +605 J+605 \text{ J}
  4. +380 J+380 \text{ J}
  5. +480 J+480 \text{ J}
Explanation: 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: ΔU=Q−W\Delta U = Q - W. For a complete cycle, the internal energy change is zero (ΔU=0\Delta U = 0), so Qnet=WnetQ_{net} = W_{net}. The net heat absorbed is Qnet=850 J−320 J=530 JQ_{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 J−75 J=455 J530 \text{ J} - 75 \text{ J} = 455 \text{ J}. Answer A (+530 J+530 \text{ J}) represents the total work without subtracting the electrical work component. Answer C (+605 J+605 \text{ J}) incorrectly adds the electrical work instead of subtracting it, suggesting a sign error in the calculation. Answer D (+380 J+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.

Question 8

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)?

  1. 0 J0 \text{ J}
  2. +1729 J+1729 \text{ J} (correct answer)
  3. −1729 J-1729 \text{ J}
  4. +3458 J+3458 \text{ J}
  5. +2494 J+2494 \text{ J}
Explanation: 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=ΔU+wq = \Delta U + w. Since both are isothermal processes with an ideal gas, ΔU=0\Delta U = 0 (internal energy depends only on temperature for ideal gases). Therefore, q=wq = w for each process. Container A (free expansion): The gas expands against vacuum, so no external pressure opposes the expansion. Thus wA=0w_A = 0, meaning qA=0q_A = 0. Container B (reversible isothermal expansion): The gas does work against external pressure during expansion. For isothermal expansion from ViV_i to Vf=2ViV_f = 2V_i: wB=nRTln⁡(VfVi)=(1.0)(8.314)(298)ln⁡(2)=1729 Jw_B = nRT \ln\left(\frac{V_f}{V_i}\right) = (1.0)(8.314)(298)\ln(2) = 1729 \text{ J} Therefore qB=1729 Jq_B = 1729 \text{ J}, and qB−qA=1729−0=1729 Jq_B - q_A = 1729 - 0 = 1729 \text{ J}. Choice A (0 J0 \text{ J}) incorrectly assumes both processes absorb the same heat. Choice C (−1729 J-1729 \text{ J}) has the sign reversed—this would be the work done by the surroundings. Choice D (+3458 J+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=wq = w since ΔU=0\Delta U = 0. Free expansion always means w=0w = 0, while reversible processes require calculating work from PP-VV relationships.

Question 9

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?

  1. q=−340 Jq = -340 \text{ J} (correct answer)
  2. q=−510 Jq = -510 \text{ J}
  3. q=−160 Jq = -160 \text{ J}
  4. q=+160 Jq = +160 \text{ J}
  5. q=−700 Jq = -700 \text{ J}
Explanation: When you encounter thermodynamics problems involving energy changes, always start with the first law of thermodynamics: ΔU=q+w\Delta U = q + w, where ΔU\Delta U is the change in internal energy, qq is heat flow, and ww is work done on the system. First, identify what you know: the internal energy decreases by 425 J, so ΔU=−425 J\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 J-180 \text{ J} to the total work. Additionally, 95 J of electrical work is done ON the system, contributing +95 J+95 \text{ J}. Therefore: w=−180+95=−85 Jw = -180 + 95 = -85 \text{ J}. Using the first law: −425=q+(−85)-425 = q + (-85), which gives q=−340 Jq = -340 \text{ J}. This matches answer A. Answer B (q=−510 Jq = -510 \text{ J}) likely results from incorrectly adding both work values as negative: −425=q+(−180−95)-425 = q + (-180 - 95). Answer C (q=−160 Jq = -160 \text{ J}) probably comes from using ΔU=+425 J\Delta U = +425 \text{ J} instead of the correct negative value. Answer D (q=+160 Jq = +160 \text{ J}) combines both errors: wrong sign for ΔU\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.

Question 10

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 Δngas=+2.0\Delta n_{gas} = +2.0 mol, what would be the enthalpy change?

  1. ΔH=−250\Delta H = -250 kJ because enthalpy change equals internal energy change
  2. ΔH=−245\Delta H = -245 kJ because ΔH=ΔU+ΔngasRT\Delta H = \Delta U + \Delta n_{gas}RT at final temperature
  3. ΔH=−255\Delta H = -255 kJ because ΔH=ΔU+ΔngasRT\Delta H = \Delta U + \Delta n_{gas}RT at initial temperature (correct answer)
  4. ΔH=−252\Delta H = -252 kJ because ΔH=ΔU+ΔngasRTˉ\Delta H = \Delta U + \Delta n_{gas}R\bar{T} using average temperature
Explanation: For the constant volume process, qV=ΔU=−250q_V = \Delta U = -250 kJ (negative because energy is released by the system). The relationship between enthalpy and internal energy is ΔH=ΔU+Δ(PV)=ΔU+ΔngasRT\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. ΔH=−250+(2.0)(8.314×10−3)(298)=−250+4.95=−255\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.

Question 11

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?

  1. ΔU=0\Delta U = 0 J because it is a cyclic process (correct answer)
  2. ΔU=+430\Delta U = +430 J because heat absorbed exceeds heat released
  3. ΔU=−430\Delta U = -430 J because work is done by the system
  4. ΔU=+850\Delta U = +850 J because internal energy depends only on temperature
Explanation: 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.

Question 12

A system undergoes a thermodynamic cycle consisting of four reversible processes: (1) isothermal compression with w1=−400w_1 = -400 J, (2) isobaric expansion with q2=+800q_2 = +800 J, (3) isothermal expansion with ΔU3=0\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?

  1. q3=+200q_3 = +200 J because net work equals net heat for cyclic processes
  2. q3=+600q_3 = +600 J because isothermal processes require heat input for expansion
  3. q3=+400q_3 = +400 J because q3=w3q_3 = w_3 for isothermal processes (correct answer)
  4. q3=+800q_3 = +800 J because heat transfer matches the isobaric expansion step
Explanation: For the complete cycle: ΔUcycle=0\Delta U_{cycle} = 0 and qnet=wnet=+200q_{net} = w_{net} = +200 J. For process 3 (isothermal): ΔU3=0\Delta U_3 = 0, so q3=w3q_3 = w_3. Process 1: ΔU1=0\Delta U_1 = 0 (isothermal), so q1=w1=−400q_1 = w_1 = -400 J. Process 2: Given q2=+800q_2 = +800 J. To find w3w_3: The net work is w1+w2+w3+w4=+200w_1 + w_2 + w_3 + w_4 = +200 J. Since qnet=q1+q2+q3+q4=+200q_{net} = q_1 + q_2 + q_3 + q_4 = +200 J, and q1=−400q_1 = -400 J, q2=+800q_2 = +800 J, we need to determine q3q_3. From the constraint that q3=w3q_3 = w_3 and the cycle relationships, q3=w3=+400q_3 = w_3 = +400 J. Choice A confuses net work with individual process heat. Choice B arbitrarily assigns a value. Choice D incorrectly assumes q3=q2q_3 = q_2.

Question 13

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 qVq_V (heat at constant volume) and qPq_P (heat at constant pressure)?

  1. qP=qV+ΔngasRTq_P = q_V + \Delta n_{gas}RT where Δngas\Delta n_{gas} accounts for gaseous product formation (correct answer)
  2. qP=qV−ΔngasRTq_P = q_V - \Delta n_{gas}RT where Δngas\Delta n_{gas} accounts for gaseous product formation
  3. qP=qVq_P = q_V because the heat released depends only on bond energies
  4. qP=qV+PVproductsq_P = q_V + PV_{products} where PVproductsPV_{products} is the pressure-volume work of products
Explanation: The relationship between constant volume and constant pressure heat is qP=qV+ΔngasRTq_P = q_V + \Delta n_{gas}RT, where Δngas\Delta n_{gas} is the change in moles of gas. This comes from ΔH=ΔU+Δ(PV)=ΔU+ΔngasRT\Delta H = \Delta U + \Delta(PV) = \Delta U + \Delta n_{gas}RT for ideal gases. Since qP=−ΔHq_P = -\Delta H and qV=−ΔUq_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 PVproductsPV_{products} rather than the change in ΔngasRT\Delta n_{gas}RT.

Question 14

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?

  1. ΔU=−850\Delta U = -850 J because the system only loses thermal energy
  2. ΔU=−1128\Delta U = -1128 J because both heat loss and expansion work decrease internal energy (correct answer)
  3. ΔU=−572\Delta U = -572 J because thermal energy loss is partially offset by expansion work
  4. ΔU=+278\Delta U = +278 J because expansion work exceeds the thermal energy released
Explanation: First, identify signs: q=−850q = -850 J (heat released by system), w=+PextΔV=1.2×101325×(2.8−0.5)×10−3=+278w = +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: ΔU=q−w=−850−278=−1128\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.

Question 15

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?

  1. ΔUtotal=+160\Delta U_{total} = +160 J because internal energy increases in first step only
  2. ΔUtotal=+650\Delta U_{total} = +650 J because ΔU=q+w\Delta U = q + w for both steps combined (correct answer)
  3. ΔUtotal=+800\Delta U_{total} = +800 J because work done on gas increases internal energy
  4. ΔUtotal=+310\Delta U_{total} = +310 J because work and heat effects are properly accounted
Explanation: Apply the first law to each step separately. Step 1: q1=+480q_1 = +480 J (absorbed), w1=−320w_1 = -320 J (work done on gas is negative work done by gas), so ΔU1=q1−w1=480−(−320)=+800\Delta U_1 = q_1 - w_1 = 480 - (-320) = +800 J. Step 2: q2=0q_2 = 0 J (adiabatic), w2=+150w_2 = +150 J (work done by gas), so ΔU2=0−150=−150\Delta U_2 = 0 - 150 = -150 J. Total: ΔUtotal=800+(−150)=+650\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.