# PS1KEYF07 - Chemistry 391 PS1 KEY 1 PV T= = nR 1.00 106 Pa...

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1 Chemistry 391 Fall 2007 PS1 KEY 1. 63 3 11 1 1.00 10 Pa 10 m 723 K 5.00 g 8.314J mol K 30.07 g mol PV T nR −− ×× == = × × 2. a) 2 2 2 2 5 33 4 5 1.00 2.016 g mol 8.314 J mol K 300 K 6.24 10 Pa 2.00 10 m 1.00 32.00 g mol 8.314 J mol K 300 K 3.90 10 Pa 2.00 10 m 6.57 10 Pa H H O O total nR T g P V T g P V P = × × = × × 2 2 22 2 2 2 mol H 1.00 2.016 mol fraction H .941 mol H mol O 1.00 2.016 1.00 32.00 mol O 1.00 32.00 mol fraction O .059 mol H mol O 1.00 2.016 1.00 32.00 1 mol fraction H = ++ = =− b) 2 2 2 2 4 4 4 1.00 g 28.02 g mol 8.314 J mol K 300 K 4.45 10 Pa 2.00 10 m 1.00 g 32.00 g mol 8.314 J mol K 300 K 3.90 10 Pa 2.00 10 m 8.35 10 Pa N N O O total T P V T P V P = × × = × × 2 2 2 2 mol N 1.00 28.02 mol fraction N .533 mol N mol O 1.00 28.02 1.00 32.00 mol O 1.00 32.00 mol fraction O .467 mol N mol O 1.00 28.02 1.00 32.00 = = 3. To obtain the Euler chain from PT VV dV dT dP TP ∂∂ ⎛⎞ =+ ⎜⎟ ⎝⎠ (1). Since the chain has a term V P T the idea is to set V constant in (1). Then

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2 0 PT VV dT dP TP ∂∂ ⎛⎞ =+ ⎜⎟ ⎝⎠ or dP dT =− or P T dP dT Therefore, now we can get a fixed V partial; namely, P VT V P Vd T V V V Td T P T P T ⎛⎞ ⎛⎞ .
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PS1KEYF07 - Chemistry 391 PS1 KEY 1 PV T= = nR 1.00 106 Pa...

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