📅 Published on: 01.12.2025
Numerical 1 – Isothermal Expansion
Question: 1 mole of an ideal gas expands isothermally at T = 300 K from volume V1 = 2 L to V2 = 5 L. Calculate the work done by the gas. Take R = 8.314 J mol-1 K-1.
Solution:
For an isothermal process (ideal gas):
W = n R T ln(V2 / V1)
Substitute values:
W = 1 × 8.314 × 300 × ln(5 / 2)
ln(5 / 2) ≈ 0.916
W ≈ 8.314 × 300 × 0.916 ≈ 2.29 × 103 J
Answer: W ≈ 2.3 × 103 J (work done by the gas).
Numerical 2 – Isothermal Compression
Question: 0.5 mole of an ideal gas is compressed isothermally at T = 400 K from V1 = 10 L to V2 = 4 L. Find the work done by the gas and on the gas.
Solution:
For an isothermal process:
W(by) = n R T ln(V2 / V1)
Substitute values:
W(by) = 0.5 × 8.314 × 400 × ln(4 / 10)
ln(0.4) ≈ -0.916
W(by) ≈ 0.5 × 8.314 × 400 × (−0.916) ≈ −1.52 × 103 J
So, W(by) ≈ −1.5 × 103 J
W(on) = −W(by) ≈ +1.5 × 103 J
Answer: Work done by the gas W(by) ≈ −1.5 kJ, work done on the gas W(on) ≈ +1.5 kJ.
Numerical 3 – Isobaric Expansion
Question: An ideal gas is heated at constant pressure P = 1.5 × 105 Pa and expands from V1 = 3 L to V2 = 8 L. Calculate the work done by the gas.
Solution:
Convert volumes to m3:
V1 = 3 L = 3 × 10−3 m3
V2 = 8 L = 8 × 10−3 m3
Change in volume:
ΔV = V2 − V1 = (8 − 3) × 10−3 = 5 × 10−3 m3
For an isobaric process:
W = P ΔV
W = 1.5 × 105 × 5 × 10−3 = 7.5 × 102 J
Answer: W = 750 J (work done by the gas).
Numerical 4 – Isobaric Heating
Question: 1 mole of an ideal gas is heated isobarically from
T1 = 300 K to T2 = 450 K. The initial volume is
V1 = 4 L. Find:
(a) the final volume V2
(b) the work done by the gas during this process.
Take R = 8.314 J mol−1 K−1.
Solution:
(a) Final volume
For an isobaric process of an ideal gas:
V2 / T2 = V1 / T1
V2 = V1 × (T2 / T1)
V2 = 4 L × (450 / 300) = 4 × 1.5 = 6 L
(b) Work done
Convert V1 to m3:
V1 = 4 × 10−3 m3
Use the ideal gas law to find pressure:
P = (n R T1) / V1
P ≈ (1 × 8.314 × 300) / (4 × 10−3) ≈ 6.24 × 105 Pa
Final volume in m3:
V2 = 6 L = 6 × 10−3 m3
Change in volume:
ΔV = V2 − V1 = (6 − 4) × 10−3 = 2 × 10−3 m3
Work done:
W = P ΔV ≈ 6.24 × 105 × 2 × 10−3 ≈ 1.25 × 103 J
Answers:
(a) V2 = 6 L
(b) W ≈ 1.25 × 103 J
Numerical 5 – Two-Step Process (Isothermal + Isobaric)
Question: An ideal gas initially at pressure
P1 = 2.0 × 105 Pa and volume
V1 = 3 L undergoes the following two steps:
1. Isothermal expansion until its volume doubles.
2. Isobaric compression at the final pressure of step 1, back to the original volume 3 L.
Calculate:
(a) Work done in the isothermal step
(b) Work done in the isobaric step
(c) Net work done over the complete cycle.
Solution:
Convert initial volume to m3:
V1 = 3 L = 3 × 10−3 m3
After isothermal expansion:
V2 = 2 V1 = 6 L = 6 × 10−3 m3
(a) Isothermal step
Work done in isothermal expansion:
W1 = P1 V1 ln(V2 / V1)
W1 = 2.0 × 105 × 3 × 10−3 × ln(2)
≈ 600 × 0.693 ≈ 4.16 × 102 J
(b) Isobaric compression
Pressure after isothermal expansion:
P2 = (P1 V1) / V2 = (2.0 × 105 × 3 × 10−3) / (6 × 10−3) = 1.0 × 105 Pa
Change in volume during compression:
ΔV = V1 − V2 = 3 × 10−3 − 6 × 10−3 = −3 × 10−3 m3
Work done:
W2 = P2 ΔV = 1.0 × 105 × (−3 × 10−3) = −3.0 × 102 J
(c) Net work
Wnet = W1 + W2 ≈ 4.16 × 102 − 3.0 × 102 ≈ 1.16 × 102 J
Answers:
(a) W1 ≈ 4.16 × 102 J (by the gas)
(b) W2 ≈ −3.0 × 102 J (on the gas)
(c) Wnet ≈ 1.16 × 102 J
Thermodynamics Numericals class 11 physics Pdf Download
Numerical 1 – Isothermal Expansion of Ideal Gas
Question: 1 mole of an ideal gas undergoes isothermal expansion at
T = 400 K from 5 L to 15 L. Calculate the work done by the gas.
(Take R = 8.314 J mol−1 K−1)Solution:
For an isothermal process for ideal gas:
W = n R T ln(V2 / V1)
Here, n = 1 mol, T = 400 K, V1 = 5 L, V2 = 15 L
W = 1 × 8.314 × 400 × ln(15 / 5)
ln(15 / 5) = ln(3) ≈ 1.099
W ≈ 8.314 × 400 × 1.099 ≈ 3.65 × 103 J
Answer: Work done by the gas ≈ 3.65 × 103 J
Numerical 2 – Heat Supplied in Isobaric Process
Question: A gas at constant pressure P = 2.2 × 105 Pa expands from 2.5 L to 6.0 L. Molar heat capacity at constant pressure Cp = 28 J mol−1 K−1. The gas contains 2 moles. Find the heat absorbed by the gas.
Solution:
Convert volume to m3:
V1 = 2.5 L = 2.5 × 10−3 m3
V2 = 6.0 L = 6.0 × 10−3 m3
ΔV = V2 − V1 = (6.0 − 2.5) × 10−3 = 3.5 × 10−3 m3
For isobaric process: P ΔV = n R ΔT ⇒ ΔT = P ΔV / (n R)
ΔT = (2.2 × 105 × 3.5 × 10−3) / (2 × 8.314)
ΔT ≈ 46.3 K
Heat absorbed, Q = n Cp ΔT
Q = 2 × 28 × 46.3 ≈ 2.59 × 103 J
Answer: Heat absorbed ≈ 2.6 × 103 J
Numerical 3 – Work Done in Isothermal Compression
Question: 0.75 mole of an ideal gas is compressed isothermally from 20 L to 8 L at 350 K. Calculate the work done on the gas. (R = 8.314 J mol−1 K−1)
Solution:
Work done by gas in isothermal process:
W(by) = n R T ln(V2 / V1)
n = 0.75, T = 350 K, V1 = 20 L, V2 = 8 L
W(by) = 0.75 × 8.314 × 350 × ln(8 / 20)
ln(8 / 20) = ln(0.4) ≈ −0.916
W(by) ≈ 0.75 × 8.314 × 350 × (−0.916) ≈ −2.0 × 103 J
Work done on gas = −W(by) ≈ +2.0 × 103 J
Answer: Work done on the gas ≈ 2.0 × 103 J
Numerical 4 – Internal Energy Change in Isochoric Heating
Question: A monoatomic ideal gas is heated at constant volume. The temperature increases from 300 K to 550 K. The gas contains 3 moles. Find the change in internal energy. (For monoatomic gas, Cv = 3R/2)
Solution:
Change in internal energy: ΔU = n Cv ΔT
ΔT = 550 − 300 = 250 K, n = 3
Cv = (3 / 2) R
ΔU = 3 × (3 / 2 × 8.314) × 250
= 3 × 1.5 × 8.314 × 250 ≈ 9.35 × 103 J
Answer: ΔU ≈ 9.35 × 103 J (increase in internal energy)
Numerical 5 – Isothermal Expansion + Isobaric Compression Cycle
Question: A gas expands isothermally from 4 L to 16 L at 300 K and then is compressed isobarically back to the original volume (4 L). Initial pressure P1 = 1.0 × 105 Pa. Find the net work done over the complete cycle.
Solution:
V1 = 4 L = 4 × 10−3 m3, V2 = 16 L = 16 × 10−3 m3
(a) Isothermal expansion (V1 → V2)
W1 = P1 V1 ln(V2 / V1)
W1 = 1.0 × 105 × 4 × 10−3 × ln(16 / 4)
= 400 × ln(4)
ln(4) ≈ 1.386
W1 ≈ 400 × 1.386 ≈ 5.55 × 102 J
(b) Isobaric compression (V2 → V1)
Using P1 V1 = P2 V2,
P2 = (P1 V1) / V2 = (1.0 × 105 × 4 × 10−3) / (16 × 10−3) = 2.5 × 104 Pa
ΔV = V1 − V2 = 4 × 10−3 − 16 × 10−3 = −12 × 10−3 m3
W2 = P2 ΔV = 2.5 × 104 × (−12 × 10−3) = −3.0 × 102 J
(c) Net work
Wnet = W1 + W2 ≈ 555 − 300 = 255 J
Answer: Net work done ≈ 2.6 × 102 J (by the gas)
Numerical 6 – Work Done in Adiabatic Expansion
Question: 2 moles of a monoatomic ideal gas undergo adiabatic expansion and its temperature falls from 500 K to 300 K. Calculate the work done by the gas. (For monoatomic gas, Cv = 3R/2, R = 8.314 J mol−1 K−1)
Solution:
For adiabatic process: Q = 0
First law: ΔU = Q − W ⇒ ΔU = −W ⇒ W = −ΔU
ΔU = n Cv ΔT
ΔT = T2 − T1 = 300 − 500 = −200 K
n = 2, Cv = (3 / 2) R
ΔU = 2 × (3 / 2 × 8.314) × (−200)
= 2 × 1.5 × 8.314 × (−200) ≈ −4.99 × 103 J
So W = −ΔU ≈ 4.99 × 103 J
Answer: Work done by the gas ≈ 5.0 × 103 J
Numerical 7 – Heat and Work in Isobaric Heating
Question: A gas is heated at constant pressure P = 1.5 × 105 Pa. Temperature increases from 290 K to 450 K. The gas contains 1.2 moles. Given: Cp = 5R/2. Find (a) work done by the gas (b) heat absorbed by the gas.
Solution:
ΔT = 450 − 290 = 160 K, n = 1.2, R = 8.314 J mol−1 K−1
(a) Work done
For isobaric process: W = n R ΔT
W = 1.2 × 8.314 × 160 ≈ 1.6 × 103 J
(b) Heat absorbed
Cp = (5 / 2) R = 2.5 × 8.314 ≈ 20.785 J mol−1 K−1
Q = n Cp ΔT = 1.2 × 20.785 × 160 ≈ 3.99 × 103 J
Answers: W ≈ 1.6 × 103 J, Q ≈ 4.0 × 103 J
Numerical 8 – Efficiency of a Heat Engine
Question: A heat engine receives 6000 J of heat and rejects 2400 J to the sink. Find its efficiency.
Solution:
Q1 = 6000 J (heat absorbed), Q2 = 2400 J (heat rejected)
Work done, W = Q1 − Q2 = 6000 − 2400 = 3600 J
Efficiency, η = W / Q1 = 3600 / 6000 = 0.6
η = 60%
Answer: Efficiency = 60%
Numerical 9 – Carnot Engine Between Two Reservoirs
Question: A Carnot engine operates between 600 K and 350 K. If it absorbs 1000 J of heat from the source, find (a) efficiency (b) work done (c) heat rejected.
Solution:
T1 = 600 K (source), T2 = 350 K (sink)
(a) Efficiency
η = 1 − (T2 / T1) = 1 − (350 / 600) = 1 − 0.5833 ≈ 0.4167
η ≈ 41.7%
(b) Work done
W = η Q1 = 0.4167 × 1000 ≈ 4.17 × 102 J
(c) Heat rejected
Q2 = Q1 − W = 1000 − 417 ≈ 583 J
Answers: η ≈ 41.7%, W ≈ 4.2 × 102 J, Q2 ≈ 5.8 × 102 J
Numerical 10 – First Law of Thermodynamics
Question: A gas absorbs 820 J of heat. During expansion, it performs 300 J of work. Then 120 J of work is done on the gas in another step. Calculate the net change in internal energy of the gas.
Solution:
Total heat absorbed: Q = +820 J
Work done by gas in first step: W1 = +300 J
Work done on gas in second step: W2 (on) = +120 J
So work done by gas in second step: W2 (by) = −120 J
Net work done by gas: W(net) = W1 + W2 (by) = 300 − 120 = 180 J
First law: ΔU = Q − W(net)
ΔU = 820 − 180 = 640 J
Answer: Net change in internal energy ΔU = +640 J
