Kinetic energy of gasses (calculations)

Question 1

A window (A = 1.2 m², R = 1.5 m²·K/W) in a room at 22°C loses heat to the outside at 8°C. What is the heat current?

Answer 1

H = A (T_in - T_out) / R = 1.2 (22 - 8) / 1.5 = 1.2 × 14 / 1.5 = 11.2 W

Question 2

A door loses heat at H = 9.0 W for 45 minutes. How much heat is lost?

Answer 2

Q = H × Δt = 9.0 × (45 × 60) = 9.0 × 2700 = 24,300 J

Question 3

A 20 m³ room at 1.013 × 10⁵ Pa and 293 K (C_P = 29 J/mol·K) loses 35,000 J. What is the temperature drop?

Answer 3

n = p V / (R T) = (1.013 × 10⁵ × 20) / (8.314 × 293) = 831 moles; ΔT = Q / (n C_P) = 35,000 / (831 × 29) = 1.45 K

Question 4

A person inhales 0.60 L of air at 1.00 atm and 25°C. How many molecules are inhaled?

Answer 4

N = p V / (k T) = (1.013 × 10⁵ × 0.60 × 10⁻³) / (1.381 × 10⁻²³ × 298) = 1.48 × 10²² molecules

Question 5

At 1500 m altitude (p₀ = 1.013 × 10⁵ Pa, T = 293 K, M = 0.0288 kg/mol), how many O₂ molecules (21%) are in 0.50 L?

Answer 5

p = p₀ e^(-g M h / R T) = 1.013 × 10⁵ e^(-(9.8 × 0.0288 × 1500) / (8.314 × 293)) = 8.47 × 10⁴ Pa; N = p V / (k T) × 0.21 = (8.47 × 10⁴ × 0.50 × 10⁻³) / (1.381 × 10⁻²³ × 293) × 0.21 = 2.20 × 10²¹

Question 6

A 2.0 L flask of ethane (M = 30.1 g/mol) at 1.013 × 10⁵ Pa and 500 K cools to 350 K. What is the final pressure?

Answer 6

p₂ = (T₂ / T₁) p₁ = (350 / 500) × 1.013 × 10⁵ = 7.09 × 10⁴ Pa

Question 7

How many grams of ethane (M = 30.1 g/mol) are in a 1.0 L flask at 1.013 × 10⁵ Pa and 450 K?

Answer 7

m = (p V M) / (R T) = (1.013 × 10⁵ × 1.0 × 10⁻³ × 0.0301) / (8.314 × 450) = 0.815 g

Question 8

What is the mass of one methane molecule (M = 16.0 g/mol)?

Answer 8

m = M / N_A = (16.0 × 10⁻³) / (6.022 × 10²³) = 2.66 × 10⁻²⁶ kg

Question 9

4 moles of gas (C_V = 25.0 J/mol·K) at 10°C are heated to 150°C. How much heat is added?

Answer 9

Q = n C_V ΔT = 4 × 25.0 × (150 - 10) = 4 × 25.0 × 140 = 14,000 J

Question 10

A 0.30 m³ container with 12 moles of gas at 1.00 × 10⁵ Pa gains 18,000 J, reaching 1.35 × 10⁵ Pa. What is C_V?

Answer 10

T₁ = p₁ V / (n R) = (1.00 × 10⁵ × 0.30) / (12 × 8.314) = 300 K; T₂ = (1.35 × 10⁵ × 0.30) / (12 × 8.314) = 405 K; C_V = Q / (n ΔT) = 18,000 / (12 × 105) = 14.29 J/mol·K

Question 11

Calculate v_rms for a gas with m = 4.00 × 10⁻²⁷ kg at 15°C.

Answer 11

v_rms = √(3 k T / m) = √(3 × 1.381 × 10⁻²³ × 288 / 4.00 × 10⁻²⁷) = √(2.98 × 10⁶) = 1726 m/s

Question 12

Gas A (m = 2.00 × 10⁻²⁷ kg) and Gas B (m = 5.00 × 10⁻²⁶ kg) are at 20°C and 1.00 × 10⁵ Pa. What is v_rms for Gas A?

Answer 12

v_rms = √(3 k T / m) = √(3 × 1.381 × 10⁻²³ × 293 / 2.00 × 10⁻²⁷) = √(6.06 × 10⁶) = 2462 m/s

Question 13

For Gas B (m = 5.00 × 10⁻²⁶ kg) at 20°C, what temperature gives v_rms = 2462 m/s?

Answer 13

v_rms² = 3 k T / m; T = v_rms² m / (3 k) = (2462² × 5.00 × 10⁻²⁶) / (3 × 1.381 × 10⁻²³) = 7.31 × 10³ K

Question 14

A window (A = 0.8 m², R = 1.3 m²·K/W) loses heat from 18°C to 5°C. What is the heat current?

Answer 14

H = A (T_in - T_out) / R = 0.8 (18 - 5) / 1.3 = 0.8 × 13 / 1.3 = 8.0 W

Question 15

A room loses 12.0 W for 2 hours. How much heat is lost?

Answer 15

Q = H × Δt = 12.0 × (2 × 3600) = 12.0 × 7200 = 86,400 J

Question 16

A 15 m³ room at 1.013 × 10⁵ Pa and 300 K (C_P = 30 J/mol·K) loses 20,000 J. What is ΔT?

Answer 16

n = p V / (R T) = (1.013 × 10⁵ × 15) / (8.314 × 300) = 609 moles; ΔT = Q / (n C_P) = 20,000 / (609 × 30) = 1.09 K

Question 17

How many N₂ molecules (28%) are in 0.70 L at 1.00 atm and 22°C?

Answer 17

N = p V / (k T) × 0.28 = (1.013 × 10⁵ × 0.70 × 10⁻³) / (1.381 × 10⁻²³ × 295) × 0.28 = 4.87 × 10²¹

Question 18

At 1000 m (p₀ = 1.013 × 10⁵ Pa, T = 288 K, M = 0.0288 kg/mol), what is the pressure?

Answer 18

p = p₀ e^(-g M h / R T) = 1.013 × 10⁵ e^(-(9.8 × 0.0288 × 1000) / (8.314 × 288)) = 8.95 × 10⁴ Pa

Question 19

A 1.2 L flask of propane (M = 44.1 g/mol) at 1.013 × 10⁵ Pa and 400 K cools to 320 K. What is p₂?

Answer 19

p₂ = (T₂ / T₁) p₁ = (320 / 400) × 1.013 × 10⁵ = 8.10 × 10⁴ Pa

Question 20

How many grams of propane (M = 44.1 g/mol) are in a 0.5 L flask at 1.013 × 10⁵ Pa and 350 K?

Answer 20

m = (p V M) / (R T) = (1.013 × 10⁵ × 0.5 × 10⁻³ × 0.0441) / (8.314 × 350) = 0.767 g

Question 21

What is the mean free path in a nebula with 40 H atoms/cm³ and r = 5.0 × 10⁻¹¹ m?

Answer 21

λ = 1 / (4 π √2 r² (N/V)) = 1 / (4 × 3.14 × 1.414 × (5.0 × 10⁻¹¹)² × 40 × 10⁶) = 5.62 × 10¹¹ m

Question 22

Calculate v_rms for H (M = 1.008 × 10⁻³ kg/mol) at 25 K

Answer 22

v_rms = √(3 R T / M) = √(3 × 8.314 × 25 / 1.008 × 10⁻³) = √(618,750) = 787 m/s

Question 23

What is the mean free time if λ = 5.0 × 10¹¹ m and v_rms = 700 m/s?

Answer 23

t = λ / v_rms = 5.0 × 10¹¹ / 700 = 7.14 × 10⁸ s

Question 24

What is the pressure in a nebula with 60 H atoms/cm³ at 18 K?

Answer 24

p = (N/V) k T = (60 × 10⁶) × (1.381 × 10⁻²³) × 18 = 1.49 × 10⁻¹⁴ Pa

Question 25

Calculate v_esc for a nebula (M = 2.5 × 10³² kg, R = 2.0 × 10¹⁶ m).

Answer 25

v_esc = √(2 G M / R) = √(2 × 6.67 × 10⁻¹¹ × 2.5 × 10³² / 2.0 × 10¹⁶) = √(833.75) = 28.9 m/s

Question 26

A 0.20 m³ tank with 8 moles at 1.00 × 10⁵ Pa gains 12,000 J, reaching 1.30 × 10⁵ Pa. What is C_V?

Answer 26

T₁ = (1.00 × 10⁵ × 0.20) / (8 × 8.314) = 300 K; T₂ = (1.30 × 10⁵ × 0.20) / (8 × 8.314) = 390 K; C_V = 12,000 / (8 × 90) = 16.67 J/mol·K

Question 27

5 moles of gas (C_V = 20.0 J/mol·K) at 5°C are heated to 100°C. How much heat is added?

Answer 27

Q = n C_V ΔT = 5 × 20.0 × (100 - 5) = 5 × 20.0 × 95 = 9500 J

Question 28

Gas A (m = 3.00 × 10⁻²⁷ kg) at 25°C has v_rms = 2000 m/s. What is T for Gas B (m = 6.00 × 10⁻²⁶ kg) to match?

Answer 28

T = v_rms² m / (3 k) = (2000² × 6.00 × 10⁻²⁶) / (3 × 1.381 × 10⁻²³) = 5.79 × 10³ K