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Q: This question requires a numeric or algebraic answer. In order to receive full c

a) T=400+273.2 =673.2 K 1 atm =1.013*10^5 pa given P=50 *1.013*10^5 =5.065*10^6 pa M=pV =>V=M/p V=nMmol/p since PV=nRT =>n=PV/RT V=(PV/RT)(Mmol/p) V=(5.065*10^6*5/8.31*673.2)(0.018/1000) V=0.0815 m^3 or 81.5 L b) T2=150+273.2 =423.2 K By ideal gas law P1V1/T1 =P2V2/T2 V2 =(P1/P2)(T2/T1)V1 V2=(50/2)(423.2/673.2)*5 V2=78.6 m^3

ID: 2104744 • Letter: T

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An electric generating plant boils water to make high-pressure steam. The steam spins a turbine that is connected to the generator. How many liters of water must be boiled to fill a 5.0-cubic-meter boiler with 50 atm of steam at 400 ° C? The steam pressure has dropped to 2.0 atm at 150 ° C as the steam exits the turbine. How much volume does it now occupy? (Points : 20)

Explanation / Answer

T=400+273.2 =673.2 K

1 atm =1.013*10^5 pa

given P=50 *1.013*10^5 =5.065*10^6 pa

M=pV =>V=M/p

V=nMmol/p

since

PV=nRT

=>n=PV/RT

V=(PV/RT)(Mmol/p)

V=(5.065*10^6*5/8.31*673.2)(0.018/1000)

V=0.0815 m^3 or 81.5 L

T2=150+273.2 =423.2 K

By ideal gas law

P1V1/T1 =P2V2/T2

V2 =(P1/P2)(T2/T1)V1

V2=(50/2)(423.2/673.2)*5

V2=78.6 m^3

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