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Thermodynamics Practice 2

# Thermodynamics Practice 2 - [Mlle/Iris” Thermodynamics...

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Unformatted text preview: [Mlle/Iris”! Thermodynamics Examnle Problems 1. A gas that is initially at a pressure of 10.0atm and having a volume of 5.0L is allowed to expand at constant temperature against a constant external pressure of 4.0atm until the new volume is 12.5L. Calculate the work done by the gas on the surroundings. 2. Calculate the work done on the system when 6.0L of a gas is compressed to 1.0L by a constant external pressure of 2.0atm. 3. A gas is allowed to expand at constant temperature ﬁ’om a volume of 10.0L to 20.0L against an external pressure of 1.0atm. Ifthe gas also absorbs 2501 of heat from the surroundings, what are the values of q, w and AB? 4. From the following equations and the enthalpy changes, a. C(gaph‘ne) + 02(g) —) (302(8) AHom = -393.5 k] b. H2(g) + 1/2 02(3) —> H20“) AHom = -285.8 k] c. 2C2H2® + 502(8) —) 4C02® + 2H20(1) AHom = -2598.8kJ calculate the standard enthalpy of formation of acetylene (Csz) from its elements: ~2C(snphi:e) + Hm) 9 Cszm 5. Verify your result ﬁ'om question 4 using standard enthalpies of formation. lPL'IIGIIIBW Thermodynamics EXIIIIIIIB PI‘IIIIIGIIIS 1. A gas that is initially at a pressure of 10.0atm and having a volume of 5.0L is allowed to expand at constant temperature against a constant external pressure of 4.0atm until the new volume is 12.5L. Calculate the work done by the gas on the surroundings. ~ 30 L4+m )( /0/, 3 J’ F/’3 : _?.0 l w “M x 0 Ti W= ’PAV \ W : [—gmmi/WIQI " 570‘) 2 2. Calculate the work done on the system when 6.0L of a gas is compressed to 1.0L by a constant external pressure of 2.0atm. w=-/’A\/ : ‘2.04+m //.0L—é.0L) : @Lm‘m: Lox/033- 3. A gas is allowed to expand at constant temperature from a volume of 10.0L to 20.0L against an external pressure of 1.0atm. If the gas also absorbs 250] of heat from the surroundings, what are the values of q, w and AE? W: (_/,'0MLW~)[20J0L —/0,0L3 : 4. From the following equations and the enthalpy changes, a. C(g'aphite) + 02(5) —> C02“) AHonm = —393.5 k] b. H2® + 1/2 02(3) —) H200) mm = -285.3 k] C. 2C2H2(g) + 502(8) —) 4C02(g) + 211200) AHom = -2598.8kJ calculate the stande enthalpy of formation of acetylene (C2H2) from its elements: \ 2C<graphhe) + H209 4 C2H2(s) ‘9 _ i— 2 ‘9 rel/urge. c) ycoz ,5) ;. lag/O ——) ZG/é (j) + 902(3) hm ~ WW”) “ ‘v ‘i W A) 1m L A v , 2H1”? " 02(7) “’9 7/7/20/4) Am” : WWW/q [/Cl‘JV‘WW'R) +V0uap -~> (/6025) AAA/Drum = 457%0/‘3 , _ “WWW?” ‘le‘L Wow/Hie) + 242mg ZCzlA/a) ANN» = +753.Z’¢f W 0"“‘04‘ 5. Verify your result from question 4 using standard enthalpies of formation. A») 2 4° C)”:- Mf": (*nﬂ’ﬂ’ﬁ) + 2(a)) 52+227£73 er»“Z?é-6l<5 ...
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Thermodynamics Practice 2 - [Mlle/Iris” Thermodynamics...

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