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7/28/2019 1st Law Worked Examples http://slidepdf.com/reader/full/1st-law-worked-examples 1/4 CHE 242: First law of Thermodynamics - Worked Examples 1. One mole of an ideal gas at 300 K is allowed to expand isothermally against a constant external pressure of 1.00 bar from a volume of 1.00 liter to a volume of 5 liters. Find the work done by the gas and the heat energy absorbed from the surroundings. Solution: Start with = 2 1 ext dV P w and see if you can obtain = w - 400 J AND q = + 400 J ;  make sure you know how to arrive here 2. Find the work necessary to compress the gas in Example 1 back to its original volume, using a constant external pressure. Is the process in Example 1 reversible? Solution: Before the expansion takes place, the pressure of the gas is: 1 P = 24.9 x 10 5 Pa = 24.9 bar (convince yourself that this is true)  Therefore, to compress the gas back to this pressure, we must use an external pressure at least 24.9 bar. Thus, we have: ( ) = = 1 2 P w ext 9960 J (convince yourself that this is true)  and the process is not reversible. 3.  Find the work done by the gas if the process in Example 1 takes place reversibly. Solution: For a reversible isothermal process involving an ideal gas, = w - 4014 J (make sure you use the correct equation) Note that the equation describing this reversible is isothermal process involves only the initial and final volumes. The work done compressing the gas back to 1  is + 4014 J. 4.  Consider an isothermal, reversible expansion of 1 mol of an ideal gas from 12.4 L at 2.4 o C to 38.1 L. Calculate q, w, and  H . Solution: = 0 and  H = 0 Isothermal change, ideal gas.

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CHE 242: First law of Thermodynamics - Worked Examples

1. One mole of an ideal gas at 300 K is allowed to expand isothermally against aconstant external pressure of 1.00 bar from a volume of 1.00 liter to a volume of 5liters. Find the work done by the gas and the heat energy absorbed from the

surroundings.

Solution:Start with ∫−=

2

1

V  ext dV Pw and see if you can obtain =w - 400 J

AND  q = + 400 J ;  make sure you know how to arrive here 

2.  Find the work necessary to compress the gas in Example 1 back to its originalvolume, using a constant external pressure. Is the process in Example 1reversible?

Solution: 

Before the expansion takes place, the pressure of the gas is:

1P = 24.9 x 105

Pa = 24.9 bar (convince yourself that this is true) 

Therefore, to compress the gas back to this pressure, we must use an external

pressure at least 24.9 bar. Thus, we have:

( ) =−−= 12 V V Pw ext  9960 J (convince yourself that this is true) 

and the process is not reversible.

3. Find the work done by the gas if the process in Example 1 takes place reversibly.

Solution:

For a reversible isothermal process involving an ideal gas,

=w - 4014 J (make sure you use the correct equation) 

Note that the equation describing this reversible is isothermal process involvesonly the initial and final volumes. The work done compressing the gas back to V 1 

is + 4014 J.

4. Consider an isothermal, reversible expansion of 1 mol of an ideal gas from 12.4 Lat 2.4oC to 38.1 L. Calculate q, w, U ∆ and  H ∆ .

Solution:

U ∆ = 0 and  H ∆ = 0 Isothermal change, ideal gas.

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Hint:

dT C nU T 

T  V ∫=∆2

1

and dT C n H T 

T  p∫=∆2

1

 

What is the value of dT for an isothermal change?

∫ ∫−=−=

2

1

2

1

V ext  PdV dV Pw ; what is the expression for P for an ideal gas?There =w -2570.2 J = - 2.57 kJ 

To find q, use the First Law of Thermodynamics, wqU  +=∆  

q = - w = 2.57 kJ, since U ∆ = 0

5. Consider an isobaric compression of 0.450 mol of an ideal gas from 22.4 L and

1.00 atm to 10.5 L. Given that C  p for the gas is  R

2

5, calculate T ∆ , q, w, U ∆ and

 H ∆ .

Solution:

First, to get T ∆ , use the ideal gas law to find the initial temperature, and then

Charles’ law (V  / T = constant) to find the final temperature.

111 nRT V P =   ⇒  ( )

( )( )314.8450.0

104.2210132500.1 311

1

==x x

nR

V PT  = 606.3 K

For isobaric (constant pressure) process according to Charles’ law

2

2

1

1

T V 

T V  =  ⇒   2

1

12 xV 

V T T  = =

3

3

104.22105.103.606−

 x x = 284.2 K

3.6062.28412 −=−=∆ T T T  = -322.1 K

For a constant pressure process, i.e. dP=0, 

( ) T nC T T nC dT nC q H   p p

T  p p ∆=−===∆ ∫ 122

1

 

= ( ) ( )322314.8

2

5450.0 −

 

 

 

  x  

= - 6690 J = - 6.69 kJ

∫ ∫−=−=2

1

2

1

V ext  PdV dV Pw , Pext  = P for reversible expansion/compression

Since P is constant

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( ) V PV V PdV PwV 

V ∆−=−−=−= ∫ 12

2

1

= ( ) 33 104.22105.1010132500.1 −−−− x x x  

= 1205 J = 1.21 kJ

Using First law of Thermodynamics

wqU  +=∆ = - 6.69 + 1.21 = -5.48 kJ

6. Consider the isochoric ( 0=∆V  ) process in which the pressure of a 2.35 mol

sample of ideal gas changes from 1.60 atm at 197 K to 2.70 atm. C V  for the gas is

 R2

3.

Solution:

First get T ∆ using the law of Gay-Lussac, P/T = constant

21

1

22

2

1

1  xPP

T T 

P

P=⇒= = 70.2

60.1

197 x = 332 K

19733212 −=−=∆ T T T  = 135 K

∫ =−=2

1

0V 

V  ext dV Pw , since dV = 0

Using First Law of thermodynamics,

wqU  +=∆ or dwdqdU  +=  

Therefore, dqdU = or qU =∆  

( ) ( ) = 

  

 =∆=−=∆= 135314.8

2

335.212 xT nC T T nC U q V V  3956 J = 3.96 kJ

Finally, PV U  H  +=   ( )PV U  H  ∆+∆=∆⇒  

PV U  H  ∆+∆=∆⇒ , for constant volume process. So to get the volume:

( )( )( )10132560.1

197314.835.2==

P

nRT V  = 2.376 x 10

-2m

3= 23.76 L

( )( ) =−+=∆ −− 33 1010132560.170.21076.2396.3 x x H  6.60 kJ

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Practice Problems

1.  State briefly the thermodynamic meaning of the terms system, open system,

closed system and isolated system.

2.  Each of the following statements is false. Either correct it or state the reason for

its being false.

(a) Heat absorbed at constant volume is called enthalpy.

(b) p

 p T 

C   

 

 

 

∂=

 

(c) H , ∆S , P and V are all state functions

(d) The compressibility factor, Z > 1 for most gases observed at high pressure is

attributable to attractive forces

(e) Cp ≈ Cv for all liquids and gases

3.  For each of the following, state whether w, q, U ∆ and  H ∆ is positive, zero, or

negative, or whether there is insufficient information to decide.

(a) Reversible isothermal expansion of a perfect gas

(b) An ideal gas undergoing a cyclic process

(c) Reversible adiabatic process

(d) Reversible heating at constant volume of an ideal gas4.  Indicate one physical condition under which each of the following formulas

holds:

(a)P

V 1

α  for gases

(b) PdV dqdU  −=  

(c)γ  PV  = Constant

(d) dU = 0 (when dV  ≠ 0)

(e) dH = C  pdT  

5.  A gas obeys the equation of state

P RT PV  α +=  

where α  is a function of T only. Starting with the definition of work done, showthat if the gas is expanded isothermally from V 1 to V 2, the reversible work will be

−=

α 

α 

2

1lnV 

V  RT w  

6.  A sample consisting of one mole of monatomic ideal gas (for which  RC v2

3= ) is

taken reversibly through the cycle

( ) ( ) ( ) ( ) ,,,,, 11222111 V P AV PC V P BV P A →→→  

where P1 = 1 atm, P2 =

2

1P , V 1 = 22.44 litres and V 2 = 2V 1, and step ( ) ( ) AC  → is

carried along an isotherm.

(a) Determine the temperature at ( ) A , ( ) B , and ( )C  .

(b) Determine (in kJ) q, w, ∆U and ∆ H for each step and for the overall cycle.