The correct mathematical expression for the first law of thermodynamics
Correct answer: C. ΔE = q - w
- A. ΔH = ΔE + PΔV
- B. q + w = ΔH
- C. ΔE = q - w
- D. ΔН +ΔЕ = w
Explanation
The first law of thermodynamics states that the total energy of a closed system remains constant. In other words, energy can neither be created nor destroyed, but it can be transformed from one form to another. This principle can be expressed mathematically as:ΔE = q + wwhere:ΔE is the change in the internal energy of the system (in Joules)q is the heat transferred to the system (in Joules)w is the work done by the system on the surroundings (in Joules)However, there is a difference between enthalpy (H) and internal energy (E):Enthalpy (H) represents the total heat content of a system, including both its internal energy and the potential energy associated with its volume and pressure.Internal energy (E) represents only the kinetic and potential energy of the particles within the system.The relation between these quantities is given by:ΔH = ΔE + PΔVwhere:P is the pressure (in Pascals)ΔV is the change in volume of the system (in cubic meters)By rearranging the enthalpy equation, we can express the change in internal energy as:ΔE = ΔH - PΔVSubstituting this into the first law of thermodynamics equation:ΔE = q + wwe get:ΔH - PΔV = q + wCombining like terms and rearranging again, we arrive at the correct expression:ΔE = q - w
Last updated
About Thermochemistry and Energetics of Chemical Reactions
Thermochemistry measures energy changes in chemical reactions and distinguishes exothermic from endothermic processes. Work covers systems, surroundings and state functions, internal energy, the first law of thermodynamics, enthalpy and Hess's law, including the sign conventions used when heat enters or leaves a system.
Practise Thermochemistry and Energetics of Chemical Reactions
728 free Thermochemistry and Energetics of Chemical Reactions MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Chemistry questions like this
Chemistry is on 12 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
1/2 H2(g) ➞ H(g) ∆H = 218 kJ mol-1.In this reaction, ∆H will be called:
1 Kcal is equal to:
2P(s) 3CI2(g) 2PCI3 ∆H= 151.8 kJ PCl3+ CI2(g) , PCI5(g); ∆H = -32.8 kJFrom the following data, the heat of formation of PCI5 comes out to be :
5 calories are equivalent to _ Joule.
5 mole of an ideal gas expand reversibly from a volume of 8 dm3 to 80 dm3 at a temperature of 27°C. Calculate the change in entropy.