A 3 2 g sample of C H X 4 gas initially at 1 0 1 . 3 2 5 kPa and 3 0 0 K was heated to 5 5 0 K . C p in SI units is
C p = 1 2 . 5 5 2 + 8 . 3 6 8 × 1 0 − 2 T
Compute Δ H in Joules.
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There's a little problem in the solution. In the question, in Cp 'T' is outside the brackets and while integrating you have used T with 8.368*10 power -2. Please review the question and solution.
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Good Problem ! . Slight More Variation Can Be Made By Giving A Relation between P And V And Then Find The Heat Capacity Of Gas
The T is outside of the parentheses.
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@Aamir Faisal Ansari check the 2nd last step in solution. You have taken T inside the parentheses. @Michelle Valdez
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Yes i think someone has unintentionally editted the problem. when i solved it T was inside the bracket only . thats why i got it correct :)
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Use Kirchoff's Law(That's essentially nothing but d H = n C p d T ).Then integrate under suitable limits ( T 1 = 3 0 0 K t o T 2 = 5 5 0 K )
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We know H = U − P V Δ H = Δ U − Δ ( P V ) At constant pressure Δ H = Δ U − P Δ V First law of thermodynamics: Q = Δ U − P Δ V Hence Δ H = Q = n ∫ T 1 T 2 C p d T where n is number of mole of methane. Since n = 1 6 3 2 = 2 mol , we have
Δ H = 2 × ∫ 3 0 0 K 5 5 0 K ( 1 2 . 5 5 2 + 8 . 3 6 8 × 1 0 − 2 T ) d T
Δ H = 2 4 0 5 8 J .