First Law of thermodynamics (questions on notes)

Question 1

What is the change in internal energy if 200 J of heat flows into a system and the system does 150 J of work?

Answer 1

The change in internal energy is 50 J (using ΔU = Q - W, where Q = 200 J and W = 150 J).

Question 2

If a system absorbs 80 J of heat along a path where it does 30 J of work, what is ΔU?

Answer 2

ΔU = 50 J (ΔU = Q - W, 80 J - 30 J = 50 J).

Question 3

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?

Answer 3

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).

Question 4

Why is ΔU the same for different paths between two states?

Answer 4

Internal energy is a state function, depending only on the initial and final states, not the path taken.

Question 5

For an ideal gas, if temperature stays constant during a process, what is ΔU?

Answer 5

ΔU = 0 J (for an ideal gas, internal energy depends only on temperature, so if T is constant, ΔU = 0).

Question 6

An ideal gas expands isothermally, doing 400 J of work. How much heat is added?

Answer 6

Q = 400 J (ΔU = 0 for isothermal, so ΔU = Q - W, 0 = Q - 400 J, Q = 400 J).

Question 7

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?

Answer 7

W = 1723 J (W = nRT ln(V2/V1), W = 1 × 8.31 × 300 × ln(2/1) ≈ 1723 J).

Question 8

In an adiabatic process, 500 J of work is done by an ideal gas. What is the change in internal energy?

Answer 8

ΔU = -500 J (Q = 0, so ΔU = Q - W, ΔU = 0 - 500 J = -500 J).

Question 9

Why is Q zero in an adiabatic process?

Answer 9

No heat flows into or out of the system during an adiabatic process.

Question 10

An ideal gas does 600 J of work in an isobaric process. If ΔU = 900 J, how much heat is added?

Answer 10

Q = 1500 J (ΔU = Q - W, 900 J = Q - 600 J, Q = 1500 J).

Question 11

What is the work done in an isochoric process where volume doesn’t change?

Answer 11

W = 0 J (work is zero because W = p ΔV, and ΔV = 0).

Question 12

If 300 J of heat is added to an ideal gas at constant volume, what is ΔU?

Answer 12

ΔU = 300 J (W = 0, so ΔU = Q - W, ΔU = 300 J - 0 = 300 J).

Question 13

For a monoatomic ideal gas, what is the molar heat capacity at constant volume?

Answer 13

CV = 3/2 R (specific to monoatomic gases like helium or neon).

Question 14

How is CP related to CV for an ideal gas?

Answer 14

CP = CV + R (where R is the gas constant, 8.31 J/mol·K).

Question 15

A monoatomic ideal gas expands adiabatically, doing 800 J of work. What is ΔU?

Answer 15

ΔU = -800 J (Q = 0, ΔU = Q - W, ΔU = 0 - 800 J = -800 J).

Question 16

In an isobaric process, a gas expands from 2 m³ to 4 m³ at 100 kPa. What is the work done?

Answer 16

W = 200,000 J (W = p ΔV, W = 100,000 Pa × (4 m³ - 2 m³) = 200,000 J).

Question 17

If ΔU = 0 for an ideal gas process and W = 250 J, does heat enter or leave the system?

Answer 17

Heat enters the system, Q = 250 J (ΔU = Q - W, 0 = Q - 250 J, Q = 250 J, positive means heat enters).

Question 18

What happens to the temperature of an ideal gas in an adiabatic expansion?

Answer 18

Temperature decreases (work is done by the gas, reducing internal energy, and thus temperature).

Question 19

A gas at constant pressure absorbs 1000 J of heat, doing 400 J of work. What is ΔU?

Answer 19

ΔU = 600 J (ΔU = Q - W, 1000 J - 400 J = 600 J).

Question 20

For 2 moles of a monoatomic gas, if ΔT = 100 K, what is ΔU?

Answer 20

ΔU = 4986 J (ΔU = n CV ΔT, CV = 3/2 R, ΔU = 2 × (3/2 × 8.31) × 100 ≈ 4986 J).

Question 21

What is the work done by a liquid expanding against a constant force of 5000 N over a volume increase of 0.01 m³?

Answer 21

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).

Question 22

If a liquid’s volume increases by 0.005 m³ under 2000 N force, how much work is done?

Answer 22

W = 10 J (W = F ΔV/A × A = 2000 N × 0.005 m³ = 10 J).

Question 23

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?

Answer 23

The work done by the gas (W = p ΔV, area = pressure × volume change).

Question 24

In a pV diagram, if a gas returns to its initial state, what is the net ΔU?

Answer 24

ΔU = 0 J (internal energy is a state function, so it’s zero for a closed cycle).

Question 25

For an ideal gas, if T increases from 300 K to 600 K at constant volume, does heat enter or leave?

Answer 25

Heat enters (ΔU = Q - W, W = 0, ΔU > 0 since T increases, so Q > 0).

Question 26

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?

Answer 26

Q = 20,000 J (Q = m cp ΔT, Q = 1 kg × 2000 J/kg·K × 10 K = 20,000 J).

Question 27

Why might cp and cV be nearly equal for a liquid like methanol?

Answer 27

Liquids are less compressible, so volume changes little, making work small and cp ≈ cV.

Question 28

In an adiabatic process, if pressure drops from 100 kPa to 80 kPa, what happens to temperature?

Answer 28

Temperature decreases (for an ideal gas, pV^γ = constant, lower pressure means lower T).

Question 29

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?

Answer 29

Q = 2306 J (W = nRT ln(V2/V1) = 1 × 8.31 × 400 × ln(2) ≈ 2306 J, ΔU = 0, so Q = W = 2306 J).

Question 30

If a gas does 1000 J of work in an isobaric expansion and Q = 2500 J, what is the final ΔU?

Answer 30

ΔU = 1500 J (ΔU = Q - W, 2500 J - 1000 J = 1500 J).