PV = nRT 101.3 ) = 1 L atm R = 8.3145 J/mol K = 0.08206 L atm/mol K AE = 9+W Asyssur q=m CAT CH20 = 4.184 3/°C 9 °C - K

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answerhappygod
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PV = nRT 101.3 ) = 1 L atm R = 8.3145 J/mol K = 0.08206 L atm/mol K AE = 9+W Asyssur q=m CAT CH20 = 4.184 3/°C 9 °C - K

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Pv Nrt 101 3 1 L Atm R 8 3145 J Mol K 0 08206 L Atm Mol K Ae 9 W Asyssur Q M Cat Ch20 4 184 3 C 9 C K 1
Pv Nrt 101 3 1 L Atm R 8 3145 J Mol K 0 08206 L Atm Mol K Ae 9 W Asyssur Q M Cat Ch20 4 184 3 C 9 C K 1 (41.62 KiB) Viewed 39 times
PV = nRT 101.3 ) = 1 L atm R = 8.3145 J/mol K = 0.08206 L atm/mol K AE = 9+W Asyssur q=m CAT CH20 = 4.184 3/°C 9 °C - K - 273 W = - Pext av Pay special attention to the signs of q, w and AH 1. A balloon is filled with 2.00 mol of an ideal gas at a constant external pressure of 1.00 atm and at a temperature of 441 K. The temperature of the gas is decreased to 290 K while the pressure remains constant at 1.00 atm. a. Calculate the work done on or by the system, w()). w() = Submit Answer: Tries 0/2 b. Calculate the heat (1) associated with this process, assuming that the heat capacity of an ideal gas is 20.8 J/mol K. 90) - Bubmit Answer: Tries 0/2 C. Calculate the change in internal energy (AE) associated with this process, in J. ΔΕ (Ο) - Submit Answer: Tries 0/2
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