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The value of ∆H for cooling 2 moles of an ideal mono atomic gas from 125°C to 25°C at constant pressure will be [given CP = \(\frac{5}{2}\) R] ……(a) -250 R (b) -500 R (c) 500 R (d) +250 R |
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Answer» (b) -500 R Ti = 125°C = 398K Tf = 25°C = 298 K ∆H = nCp (Tf – Ti ) ∆H = 2 x \(\frac{5}{2}\) R(298 – 398) ∆H = -500 R |
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