First Law of thermodynamics (calculations) Flashcards
A system (not an ideal gas) goes from state A to state B with 120 J of heat flowing in and 50 J of work done by the system. What is the change in internal energy?
ΔU = 70J
A system (not an ideal gas) moves from state A to state B. Along one path, 80 J of heat flows in and 30 J of work is done. Along another path, 20 J of work is done. How much heat flows in on the second path?
50 J of heat flows in (ΔU = Q - W, ΔU = 80 J - 30 J = 50 J, then 50 J = Q - 20 J, Q = 50 J + 20 J = 70 J).
A system starts at state A with internal energy 0 J, moves to state C with internal energy 15 J, and does 10 J of work. How much heat is absorbed?
25 J of heat is absorbed (ΔU = UC - UA = 15 J - 0 J = 15 J, ΔU = Q - W, 15 J = Q - 10 J, Q = 25 J).
A system moves from state C (internal energy 15 J) to state B (internal energy 25 J) with no work done. What is the heat absorbed?
10 J of heat is absorbed (ΔU = UB - UC = 25 J - 15 J = 10 J, ΔU = Q - W, 10 J = Q - 0 J, Q = 10 J).
1 mol of a monoatomic ideal gas (CV = 3/2 R, R = 8.31 J/mol·K) expands isobarically from 1 m³ to 3 m³ at 100 kPa. What is the work done?
The work done is 200,000 J (W = p ΔV, W = 100,000 Pa × (3 m³ - 1 m³) = 200,000 J).
0.5 mol of an ideal gas expands isothermally from 1 m³ to 3 m³ at 250 K, doing 2282 J of work. How much heat enters the system?
2282 J of heat enters the system (ΔU = Q - W, 0 = Q - 2282 J, Q = 2282 J).
0.5 mol of an ideal gas expands isothermally from 1 m³ to 3 m³ at 250 K (R = 8.31 J/mol·K). What is the work done by the gas?
The work done is 2282 J (W = nRT ln(V2/V1), W = 0.5 × 8.31 × 250 × ln(3/1) ≈ 2282 J).
0.5 mol of an ideal gas expands isothermally at 250 K from 1 m³ to 3 m³. What is the change in internal energy?
The change in internal energy is 0 J (isothermal process for an ideal gas, ΔU = 0).
1 mol of a monoatomic ideal gas (CV = 3/2 R, R = 8.31 J/mol·K) expands isothermally from 2 m³ to 4 m³, doing 2000 J of work. What is the heat added?
2000 J of heat is added (ΔU = 0 for isothermal, ΔU = Q - W, 0 = Q - 2000 J, Q = 2000 J).
1 mol of a monoatomic ideal gas (CV = 3/2 R, R = 8.31 J/mol·K) expands isobarically from 1 m³ to 3 m³ with ΔT = 24,066 K. What is ΔU?
ΔU = 300,000 J (ΔU = nCVΔT, ΔU = 1 × (3/2 × 8.31) × 24,066 ≈ 300,000 J).
1 mol of a monoatomic ideal gas expands isothermally from 2 m³ to 4 m³, doing 2000 J of work (R = 8.31 J/mol·K). What is the temperature?
The temperature is 347 K (W = nRT ln(V2/V1), 2000 = 1 × 8.31 × T × ln(4/2), T ≈ 347 K).
1 mol of a monoatomic ideal gas (CV = 3/2 R) expands adiabatically from 1 m³ to 2 m³, doing 1800 J of work. What is the change in internal energy?
The change in internal energy is -1800 J (Q = 0, ΔU = Q - W, ΔU = 0 - 1800 J = -1800 J).
1 mol of a monoatomic ideal gas (CV = 3/2 R, R = 8.31 J/mol·K) expands isobarically from 1 m³ to 3 m³, doing 200,000 J of work. What is the temperature change?
The temperature change is 24,066 K (W = nRΔT, 200,000 = 1 × 8.31 × ΔT, ΔT ≈ 24,066 K).
1 mol of a monoatomic ideal gas expands isobarically from 1 m³ to 3 m³ with 200,000 J of work and ΔU = 300,000 J. What is the heat added?
500,000 J of heat is added (ΔU = Q - W, 300,000 = Q - 200,000, Q = 500,000 J).
2 mol of N2 gas (CV = 20.76 J/mol·K) is heated at constant volume from 300 K to 500 K. What is the heat added?
8312 J of heat is added (Q = nCVΔT, ΔT = 500 - 300 = 200 K, Q = 2 × 20.76 × 200 = 8312 J).
2 mol of N2 gas (CV = 20.76 J/mol·K) is heated at constant volume from 300 K with 8312 J of heat. What is the final temperature?
The final temperature is 500 K (Q = nCVΔT, 8312 = 2 × 20.76 × ΔT, ΔT = 200 K, T2 = 300 + 200 = 500 K).
2 mol of N2 gas expands isobarically from 1 m³ to 2 m³ starting at 500 K (R = 8.31 J/mol·K). What is the final temperature?
The final temperature is 1000 K (V2/V1 = T2/T1, 2/1 = T2/500, T2 = 1000 K).
2 mol of N2 gas (CV = 20.76 J/mol·K) goes from 300 K to 1000 K over two steps. What is the total ΔU?
The total ΔU is 29,064 J (ΔU = nCVΔT, ΔT = 1000 - 300 = 700 K, ΔU = 2 × 20.76 × 700 = 29,064 J).
Methanol expands against a force of 4.0 × 10^4 N with a volume increase of 5.4 × 10^-4 m³. What is the work done?
The work done is 2160 J (W = F ΔV/A × A = 4.0 × 10^4 × 5.4 × 10^-4 = 2160 J).
Methanol (0.015 m³, β = 1.20 × 10^-3 K^-1) expands from 15°C to 45°C. What is the increase in volume?
The increase in volume is 5.4 × 10^-4 m³ (ΔV = V0 β ΔT, ΔV = 0.015 × 1.20 × 10^-3 × 30 = 5.4 × 10^-4 m³).
2 mol of N2 gas expands isobarically from 1 m³ to 2 m³ from 500 K to 1000 K (R = 8.31 J/mol·K). What is the work done?
The work done is 8310 J (W = nRΔT, ΔT = 1000 - 500 = 500 K, W = 2 × 8.31 × 500 = 8310 J).
2 mol of N2 gas (CP = 29.07 J/mol·K) expands isobarically from 500 K to 1000 K. What is the heat added?
29,070 J of heat is added (Q = nCPΔT, ΔT = 1000 - 500 = 500 K, Q = 2 × 29.07 × 500 = 29,070 J).
Methanol (mass 11.865 kg, cp = 2.51 × 10^3 J/kg·K) is heated from 15°C to 45°C. What is the heat added?
8.94 × 10^5 J of heat is added (Q = m cp ΔT, Q = 11.865 × 2.51 × 10^3 × 30 = 8.94 × 10^5 J).
Methanol is heated from 15°C to 45°C with Q = 8.94 × 10^5 J and W = 2160 J. What is the change in internal energy?
The change in internal energy is 8.92 × 10^5 J (ΔU = Q - W, 8.94 × 10^5 - 2160 = 8.92 × 10^5 J).
3 mol of helium (CV = 3/2 R, R = 8.31 J/mol·K) goes from 400 K to 800 K at constant volume. What is the change in internal energy?
The change in internal energy is 14,958 J (ΔU = nCVΔT, ΔT = 800 - 400 = 400 K, ΔU = 3 × (3/2 × 8.31) × 400 = 14,958 J).
For methanol heated with Q = 8.94 × 10^5 J, W = 2160 J, and ΔU = 8.92 × 10^5 J, why is cp close to cV?
The small work (W) compared to Q means volume change is minimal, so cp ≈ cV.
3 mol of helium (CV = 3/2 R, R = 8.31 J/mol·K) goes isobarically from 800 K to 400 K. What is the change in internal energy?
The change in internal energy is -14,958 J (ΔU = nCVΔT, ΔT = 400 - 800 = -400 K, ΔU = 3 × (3/2 × 8.31) × -400 = -14,958 J).
3 mol of helium expands isothermally at 400 K, doing 9000 J of work (R = 8.31 J/mol·K). What is the heat added?
9000 J of heat is added (ΔU = 0, ΔU = Q - W, 0 = Q - 9000, Q = 9000 J).
Air rises adiabatically from sea level (20°C, 1.013 × 10^5 Pa) to 900 m (0.91 × 10^5 Pa) with γ = 1.41. What is the final temperature?
The final temperature is 291 K (p/p0 = (T/T0)^(γ/(γ-1)), 0.91/1.013 = (T/293)^(1.41/0.41), T ≈ 291 K).
Air at 20°C and 60% humidity (dew point 9°C) rises adiabatically to 291 K. Does it form clouds?
No, it doesn’t form clouds yet (291 K = 18°C > 9°C, temperature must drop below dew point to condense).
