GATE MT · 2026 · MT
Question 56: Laws of thermodynamics
Give it a try, then reveal the answer to check your work.
Question 56
MTNAT2 marksNeeds review
With reference to the Ellingham diagram, the standard Gibbs free energy change for the oxidation of solid metal M (s) and liquid metal M (l) is given below:
Reaction I:
M (s) + O₂ (g) → MO₂ (s)
ΔG° = (− 338900 − 15.2T lnT + 247T) Joules;
from T = 300 K to melting point.
Reaction II:
M (l) + O₂ (g) → MO₂ (s)
ΔG° = (− 390800 − 15.2T lnT + 285.3T) Joules;
from melting point to T = 1800 K
The melting point of the metal M (rounded off to one decimal place) is _____________ Kelvin.
Log in to check the answer
You can attempt this question now. Sign in to view the answer key, explanations and discussion.Log in to check answerEXPLORE & PRACTISE
Paper and topic context
See where this question comes from and choose what to practise next.
2026MT
GATE MT 2026 · Paper 1
Metallurgical Engineering · Original paper archive
- Section
- Metallurgical Engineering
- Question type
- Numerical answer (NAT)
- Marks
- 2 marks
Practise related topics
Explore more previous year questions using this question’s imported topic tags.
metallurgy
GATE MT previous year questions
materials science
GATE MT previous year questions
materials engineering
GATE MT previous year questions
laws thermodynamics
GATE MT previous year questions
Laws of thermodynamics
GATE MT previous year questions
Comments
Ask a question, share your approach, or help someone understand.
Log in to view the discussion
You’re not logged in. Log in or create an account to see discussions and add comments.