Problem Solving In Chm02 Notes.docx
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CO2
CO3 A. WORK DONE IF IT EXPANDS IN A VACUUM TAKE NOTE: VACUUM = (P=0) W = - (0)(6L -2L) =0 B. AGAINST A VACUUM P= 1.2 ATM W = - (1.20 ATM) (6L-2L) = -4.8 L/ATM 1L/ATM = 103. J COVERT LXATM TO J -4.8 L/ATM * 103 J/ LXATM = -486. 24 J
2. CHANGE IN E =? W = 462 J q= -128 J TRIANGLE E = q+W = -128 J + 462 J = 334 J
CO3 A. WORK DONE IF IT EXPANDS IN A VACUUM TAKE NOTE: VACUUM = (P=0) W = - (0)(6L -2L) =0 B. AGAINST A VACUUM P= 1.2 ATM W = - (1.20 ATM) (6L-2L) = -4.8 L/ATM 1L/ATM = 103. J COVERT LXATM TO J -4.8 L/ATM * 103 J/ LXATM = -486. 24 J
2. CHANGE IN E =? W = 462 J q= -128 J TRIANGLE E = q+W = -128 J + 462 J = 334 J
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