Explain why need modifications about temperature and keep pressure atom
1 answer:
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Answer:
1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Explanation:
<u>1) To show that heat flows spontaneously from high temperature to low temperature </u>
example :
Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )
Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment
Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )
Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
<u>2) Entropy change for Decomposition of mercuric oxide </u>
2HgO (s) → 2Hg(l) + O₂ (g)
Δs = positive
there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C
hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =
Δh of reaction = 181.6 KJ
Temp = 500 + 273 = 773 k
hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
