First Law of thermodynamics (questions on notes) Flashcards
What is the change in internal energy if 200 J of heat flows into a system and the system does 150 J of work?
The change in internal energy is 50 J (using ΔU = Q - W, where Q = 200 J and W = 150 J).
If a system absorbs 80 J of heat along a path where it does 30 J of work, what is ΔU?
ΔU = 50 J (ΔU = Q - W, 80 J - 30 J = 50 J).
A system follows two paths from state X to Y. Path A has Q = 120 J and W = 40 J. If ΔU is the same for both paths, what is Q for Path B where W = 60 J?
Q = 100 J (ΔU = Q - W, ΔU = 120 J - 40 J = 80 J, so for Path B: 80 J = Q - 60 J, Q = 140 J - 60 J = 100 J).
Why is ΔU the same for different paths between two states?
Internal energy is a state function, depending only on the initial and final states, not the path taken.
For an ideal gas, if temperature stays constant during a process, what is ΔU?
ΔU = 0 J (for an ideal gas, internal energy depends only on temperature, so if T is constant, ΔU = 0).
An ideal gas expands isothermally, doing 400 J of work. How much heat is added?
Q = 400 J (ΔU = 0 for isothermal, so ΔU = Q - W, 0 = Q - 400 J, Q = 400 J).
What is the work done by an ideal gas expanding from 1 m³ to 2 m³ isothermally at 300 K with n = 1 mol and R = 8.31 J/mol·K?
W = 1723 J (W = nRT ln(V2/V1), W = 1 × 8.31 × 300 × ln(2/1) ≈ 1723 J).
In an adiabatic process, 500 J of work is done by an ideal gas. What is the change in internal energy?
ΔU = -500 J (Q = 0, so ΔU = Q - W, ΔU = 0 - 500 J = -500 J).
Why is Q zero in an adiabatic process?
No heat flows into or out of the system during an adiabatic process.
An ideal gas does 600 J of work in an isobaric process. If ΔU = 900 J, how much heat is added?
Q = 1500 J (ΔU = Q - W, 900 J = Q - 600 J, Q = 1500 J).
What is the work done in an isochoric process where volume doesn’t change?
W = 0 J (work is zero because W = p ΔV, and ΔV = 0).
If 300 J of heat is added to an ideal gas at constant volume, what is ΔU?
ΔU = 300 J (W = 0, so ΔU = Q - W, ΔU = 300 J - 0 = 300 J).
For a monoatomic ideal gas, what is the molar heat capacity at constant volume?
CV = 3/2 R (specific to monoatomic gases like helium or neon).
How is CP related to CV for an ideal gas?
CP = CV + R (where R is the gas constant, 8.31 J/mol·K).
A monoatomic ideal gas expands adiabatically, doing 800 J of work. What is ΔU?
ΔU = -800 J (Q = 0, ΔU = Q - W, ΔU = 0 - 800 J = -800 J).
In an isobaric process, a gas expands from 2 m³ to 4 m³ at 100 kPa. What is the work done?
W = 200,000 J (W = p ΔV, W = 100,000 Pa × (4 m³ - 2 m³) = 200,000 J).
If ΔU = 0 for an ideal gas process and W = 250 J, does heat enter or leave the system?
Heat enters the system, Q = 250 J (ΔU = Q - W, 0 = Q - 250 J, Q = 250 J, positive means heat enters).
What happens to the temperature of an ideal gas in an adiabatic expansion?
Temperature decreases (work is done by the gas, reducing internal energy, and thus temperature).
A gas at constant pressure absorbs 1000 J of heat, doing 400 J of work. What is ΔU?
ΔU = 600 J (ΔU = Q - W, 1000 J - 400 J = 600 J).
For 2 moles of a monoatomic gas, if ΔT = 100 K, what is ΔU?
ΔU = 4986 J (ΔU = n CV ΔT, CV = 3/2 R, ΔU = 2 × (3/2 × 8.31) × 100 ≈ 4986 J).
What is the work done by a liquid expanding against a constant force of 5000 N over a volume increase of 0.01 m³?
W = 50 J (W = F Δx, but W = p ΔV, p = F/A, here W = F ΔV/A × A = 5000 N × 0.01 m³ = 50 J).
If a liquid’s volume increases by 0.005 m³ under 2000 N force, how much work is done?
W = 10 J (W = F ΔV/A × A = 2000 N × 0.005 m³ = 10 J).
A pV diagram shows an ideal gas going from 1 m³ to 3 m³ at constant pressure. What does the area under the curve represent?
The work done by the gas (W = p ΔV, area = pressure × volume change).
In a pV diagram, if a gas returns to its initial state, what is the net ΔU?
ΔU = 0 J (internal energy is a state function, so it’s zero for a closed cycle).
For an ideal gas, if T increases from 300 K to 600 K at constant volume, does heat enter or leave?
Heat enters (ΔU = Q - W, W = 0, ΔU > 0 since T increases, so Q > 0).
What is the heat added to 1 kg of a liquid with cp = 2000 J/kg·K when T rises from 20°C to 30°C?
Q = 20,000 J (Q = m cp ΔT, Q = 1 kg × 2000 J/kg·K × 10 K = 20,000 J).
Why might cp and cV be nearly equal for a liquid like methanol?
Liquids are less compressible, so volume changes little, making work small and cp ≈ cV.
In an adiabatic process, if pressure drops from 100 kPa to 80 kPa, what happens to temperature?
Temperature decreases (for an ideal gas, pV^γ = constant, lower pressure means lower T).
A gas expands from 1 m³ to 2 m³ isothermally at 400 K (n = 1 mol, R = 8.31 J/mol·K). What is Q?
Q = 2306 J (W = nRT ln(V2/V1) = 1 × 8.31 × 400 × ln(2) ≈ 2306 J, ΔU = 0, so Q = W = 2306 J).
If a gas does 1000 J of work in an isobaric expansion and Q = 2500 J, what is the final ΔU?
ΔU = 1500 J (ΔU = Q - W, 2500 J - 1000 J = 1500 J).
