Ideal Gas Solutions

8/2/2019 Ideal Gas Solutions

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Spring 2011

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ME 201Thermodynamics

Ideal Gas Practice Problems Solutions

1. Determine the entropy change for air as it goes from 285 K and 150 kPa to 1850 K and 1000kPa.

Solution:Our entropy change will be given by

)P/Pln(Rssss 12o1

o212

So we go to the air table (A-17, pg 934) and fill in our table belowSubstance Type: Ideal Gas (air)

Process: UnknownState 1 State 2T1 = 285 K T2 = 1850 K

P1 = 150 kPa P2 = 1000 kPas

o1 = 1.65055 kJ/(kgK) so2 =3.7023 kJ/(kgK)

Italicizedvalues read from air tables or calculated from ideal gas equation.Now calculating

)P/Pln(Rssss 12o1

o212 = 3.7023 - 1.65055(0.287)ln(1000/150)

= 1.5073 kJ/(kgK)

2. Determine the internal energy change for air as it undergoes an isometric process from 320 Kand 72 kPa to 720 kPa.

Solution:Our internal energy change will be given by, u

2u

1, where the us come form the tables

So we go to the air table (A-17, pg 934) and fill in our table belowSubstance Type: Ideal Gas (air)

Process: UnknownState 1 State 2T1 = 320 K T2 = 1967 KP1 = 72 kPa P2 = 720 kPau1 = 228.42 kJ/kg u2 = 1646.95 kJ/kg

Italicizedvalues read from air tables.

Bold values are calculated.We note that our second state is not fixed (we only know the pressure), but we know the process,so that

v2 = v1From our ideal gas law we have

kg/m2756.172

)320)(287.0(

P

RTv

3

1

11

andv2 = 1.2756 m

3/kg


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ME 201 Thermodynamics Spring 2011

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We have now fixed our second state and can calculate the temperature using the ideal gas law

K1967287.0

)2756.1)(720(

R

vPT 222

We can now go to the air table and use interpolation to findu2 = 1646.95 kJ/kg

Thenu = 1646.95 - 228.42 = 1418.53 kJ/kg

3. Determine the enthalpy change (in kJ/kg) for OH as is goes from 2400 K and 1300 kPa to1600 K and 700 kPa.

Solution:Our enthalpy change will be given by, h2h1, where the hs come form the tablesSo we go to the air table (A-25, pg 947) and fill in our table below

Substance Type: Ideal Gas (air)Process: Unknown

State 1 State 2

T1 =2400 K T2 =1600 KP1 = 1300 kPa P2 = 700 kPah1 = 77,015 kJ/kmole h2 =49,358 kJ/kmole

Italicizedvalues read from air tables or calculated from ideal gas equation.Then our change in enthalpy is

h = 49,358 - 77,015 = -27,657 kJ/kmoleBut we want this in kJ/kg, so we must divide by the molecular of OH, which is 17 kg/kmole, sothat

h = -27,657/17 = -1627 kJ/kmole

4. Determine the internal energy change (in kJ/kg) for OH as it undergoes an isentropic process

from 3200 K and 7.2 MPa to 720 kPa.Solution:Our internal energy change will be given by, u2u1, where the us come form the tablesSo we go to the air table (A-25, pg 947) and fill in our table below

Substance Type: Ideal Gas (air)Process: Unknown

State 1 State 2T1 = 3200 K T2 = KP1 = 7200 kPa P2 = 720 kPa

1u = 79,539 kJ/kmol 2u = 43,004 kJ/kmolItalicizedvalues read from air tables.

Bold values are calculated.We note that our second state is not fixed (we only know the pressure), but we know the process,so that

0)P/Pln(Rssss 12uo1

o212

We know everything in this equation, except for o2s , which we can solve for

)P/Pln(Rss 12uo1

o2


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ME 201 Thermodynamics Spring 2011

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From the OH tables we find

K)kJ/(kmol259.093so1

Then calculating

K)kJ/(kmol239.95)7200/720ln()314.8(259.093so2

This now fixes our second state, which allows us to go to the OH tables and interpolate to find2u = 43,004 kJ/(kmolK)

Then our change in internal energy is

u = 43,004 - 79,539 = -36,535 kJ/kmole

But we want this in kJ/kg, so we must divide by the molecular of OH, which is 17 kg/kmole, sothat

h = -36,535/17 = -2149 kJ/kmole

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Ideal Gas Solutions