---
title: "Hydrogen in rigid vessel: heat, entropy change"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:hydrogen-rigid-vessel-entropy-produced"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/hydrogen-rigid-vessel-entropy-produced
site: BCT Engineering
---

# Hydrogen in rigid vessel: heat, entropy change

Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics

- Asked in: 2072 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/hydrogen-rigid-vessel-entropy-produced
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

A rigid vessel consist of 0.4 kg of hydrogen initially at 200 kPa and 27°C, heat is transferred to the system from a reservoir at 600K until its temperature reaches at 450 K. Determine heat transfer, the change in entropy of hydrogen and the amount of entropy produced. (Take cv = 10.183 J/KgK)

## Short answer

At constant volume, Q = m cv ΔT = 0.4×10.183×150 = 611 kJ. ΔS_H₂ = m cv ln(T₂/T₁) = 1.652 kJ/K. Reservoir entropy change = −Q/T_R = −1.018 kJ/K. Entropy produced S_gen = 0.633 kJ/K.

## Answer

**Reading of the data.** cv = 10.183 is given in J/kg·K, but for hydrogen cv ≈ 10.183 **kJ/kg·K**, so it is used in kJ/kg·K.

**Given:** m = 0.4 kg, P₁ = 200 kPa, T₁ = 27 + 273 = 300 K, T₂ = 450 K, reservoir at T_R = 600 K, rigid vessel (V = constant, W = 0).

#### Heat transfer
First law with W = 0: Q = ΔU = m cv (T₂ − T₁)

Q = 0.4 × 10.183 × (450 − 300) = **611.0 kJ** (into the hydrogen)

#### Entropy change of hydrogen
Constant volume: ΔS_H₂ = m cv ln(T₂/T₁)

ΔS_H₂ = 0.4 × 10.183 × ln(450/300) = 4.0732 × 0.40547 = **1.6515 kJ/K**

(The final pressure is P₂ = P₁T₂/T₁ = 300 kPa; it is not needed for ΔS here.)

#### Entropy change of the reservoir
The reservoir loses Q at constant 600 K:

ΔS_res = −Q/T_R = −611.0/600 = -1.0183 kJ/K

#### Entropy produced
S_gen = ΔS_H₂ + ΔS_res = 1.6515 − 1.0183

**Answer: Q = 611 kJ, ΔS_H₂ = 1.652 kJ/K, S_gen = 0.633 kJ/K.** S_gen > 0, so heat transfer across the finite temperature difference is irreversible.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the rigid vessel, heat transfer is:
   A. Q = m cp ΔT
   B. Q = m cv ΔT, since W = 0
   C. Q = m R ΔT
   D. Q = 0
   Answer: B. With V constant, W = 0 and Q = ΔU = m cv (T2 − T1) = 611 kJ. (Key point: Heat in rigid vessel)

2. ΔS of hydrogen at constant volume is:
   A. m cv ln(T2/T1)
   B. m cv ln(P2/P1) − R
   C. Q/T_R
   D. m cp ln(T2/T1)
   Answer: A. At constant volume ds = cv dT/T, giving 1.65 kJ/K. (Key point: Entropy change of hydrogen)

3. The reservoir at 600 K supplying 611 kJ has entropy change:
   A. +1.018 kJ/K
   B. −1.018 kJ/K
   C. −611 kJ/K
   D. 0
   Answer: B. It loses heat at constant temperature: −Q/T_R = −611/600. (Key point: Reservoir entropy)

4. Entropy produced is:
   A. 1.652 + 1.018 kJ/K
   B. 1.018 kJ/K
   C. ΔS_H2 + ΔS_res = 0.633 kJ/K
   D. 0
   Answer: C. Sum of system and reservoir changes; positive because of heat transfer across a finite temperature difference. (Key point: Entropy produced)

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