StudentAidUnit 6: Heat and Thermodynamics — Numericals
11th Class Physics · Unit 6: Heat and Thermodynamics
6.1.A gas occupies 6.0 L of volume at a pressure of 12 atm. What will be the volume of gas if the pressure is increased by 2.0 atm, assuming that temperature remains constant.
Given
Initial volume
V1 = 6 L
Initial pressure
P1 = 12 atm
Final pressure
P2 = 14 atm
Final volume
V2 = ?
Formula
Boyle's Law (constant temperature)
P1V1 = P2V2
Working
Solving for V₂
V2 = P1V1P2
V2 = 12 × 614
Result
Final volume
V2 = 5.14 L
6.2.In a vacuum chamber which is connected to a cryogenic pump, pressure is as low as 1.00 nPa is being attained calculate the number of molecules in 1 cm³ vessel at their pressure of and temperature of 300K.
6.3.A gas undergoes a thermodynamic process where it absorbs 500 J of heat energy and performs 300 J work on its surroundings. Calculate the change in internal energy of the gas.
Given
Heat absorbed
Q = 500 J
Work done by the system
w = 300 J
Change in internal energy
Δu = ?
Formula
First law of thermodynamics
Q = Δu + w
Working
Solving for ΔU
Δu = Q - W
Δu = 500 - 300
Result
Change in internal energy
Δu = 200 J
6.4.A Carnot engine is operating between a high temperature reservoir at 600 K and a low temperature reservoir at 300 K. Calculate: i. The maximum possible efficiency ii. The amount of work output if the engine absorbs 500 J of heat from the high temperature reservoir.
Given
High temperature reservoir
Th = 600 K
Low temperature reservoir
Tc = 300 K
Heat absorbed from hot reservoir
Qh = 500 J
Maximum efficiency
%eta = ?
Work output
W = ?
Formula
Carnot efficiency formula (part i)
%eta = left(1 - TcThright) × 100%
Working
%eta = left(1 - 300600right) × 100%
= (1 - 0.5) × 100%
Result
Maximum efficiency
%eta = 50%
Formula
Efficiency definition (part ii)
eta = WQh
Working
W = eta × Qh
W = 50100 × 500
Result
Work output
W = 250 J
6.5.A refrigerator extracts 1200 J of heat from its interior (the cold reservoir) and releases 1800 J of heat to the surrounding environment (the hot reservoir) during each cycle. Calculate: i. the work input required per cycle. ii. the co-efficient of performance (E) of the refrigerator.
Given
Heat extracted from cold reservoir
QC = 1200 J
Heat released to hot reservoir
QH = 1800 J
Work input required per cycle
W = ?
Co-efficient of performance
E = ?
Formula
Energy balance for refrigerator (part i)
W = QH - QC
W = 1800 - 1200
Result
Work input
W = 600 J
Formula
Co-efficient of performance definition (part ii)
E = QCW
E = 1200600
Result
Co-efficient of performance
E = 2
6.6.Calculate the entropy change when 1.0 mole of ice at 0°C melts to form liquid water at the same temperature.
Given
Number of moles
n = 1 mole
Temperature
T = 0°C = 273 K
Latent heat of fusion for ice
Lf = 6000 J/mol
Change in entropy
Δs = ?
Formula
Entropy change formula
Δs = QT
Working
Heat absorbed during fusion
Q = n Lf = 1 × 6000 = 6000 J
Putting Q in Eq. (1)
Δs = 6000273
Result
Change in entropy
Δs = 22 Jk-1
6.7.A gas occupies 400 ml at 20°C. What volume will it occupy at 80°C, assuming constant pressure?
Given
Initial volume
V1 = 400 mL
Initial temperature
T1 = 20°C = 20 + 273 = 293 K
Final temperature
T2 = 80°C = 80 + 273 = 353 K
Final volume
V2 = ?
Formula
Charles's Law (constant pressure)
V1T1 = V2T2
Working
Solving for V₂
V2 = V1 × T2T1
V2 = 400293 × 353
Result
Final volume
v2 = 482 mL
6.8.A gas has a pressure of 2 atm at 300 K. What pressure will it have at 450 K, assuming constant volume?
Given
Initial pressure
P1 = 2 atm
Initial temperature
T1 = 300 K
Final temperature
T2 = 450 K
Final pressure
P2 = ?
Formula
Gay-Lussac's Law (constant volume)
P1T1 = P2T2
Working
Solving for P₂
P2 = P1T1 × T2
P2 = 2300 × 450
Result
Final pressure
P2 = 3 atm