ABLE 9.2 Thermodynamic Properties of an Ideal Gas Expressiorn Application Monato

Question

ABLE 9.2 Thermodynamic Properties of an Ideal Gas Expressiorn Application Monatomic ideal gas Polyatomic ideal gas (value must be measured experimentally) All ideal gases Monatomic ideal gas Go-iR =1R + R mic el a (pei ri de e of Cy) All ideal gases All ideal gases Example 9.2 Consider 2.00 mol of a monatomic ideal gas that is taken from state A (PA = 2.00 atm, VA= 10.0 L) to state B (PB= 1.00 atm, VB-30.0 L) by two lifferent pathways: Vc = 30.0 L (Pc = 2.00 atm State A VA 10.0I PA= 2.00 atmn). State B VB 30.0 L 4PB 1.00 atm VD 10.0I PD 1.00 atm Calculate q, w, ??, and ?? for both pathways. Solution

Explanation / Answer

From state A to stare B via upper path given in the question

Step 1

At constant pressure (isobaric process) = 2 atm

Volume change = 30 L - 10 L = 20 L

Work done = - P(V2 - V1) = - 2 atm x 20 L

= - 40 L-atm x 101.325 J/L-atm

W = - 4053 J

For isobaric process

W = - nR (T2 - T1)

Change in temperature (T2 - T1)

= 4053 J / (2 mol x 8.314 J/mol·K)

= 243.7 K

At constant pressure

q 1 = n x Cp x ? T = ?H1

= (2mol) x (5/2R) x (243.7 K)

q1 = ?H1 = 10132.5 J

? E1 = n x Cv x ? T

= (2mol) x (3/2R) x (243.7 K)

= 6079.5 J

Step 2

At constant volume (isochoric process)

Volume = 30 L

Pressure changes ?P = 1 atm - 2atm = - 1 atm

Change in temperature

?T = ?PV/nR

= - 1atm x 30 L / (2 mol x 0.0821 L-atm/mol-K )

= - 182.7 K

q2 = n x C v x ? T = ?E2

= (2 mol) x (3/2R) x (-182.7 K)

q2 =?E2 = - 4.6 x 10 ^3 J

?H2 = n x Cp x ?T

= (2 mol) x (5/2R) x (-182.7 K)

= - 7.6 x 10^3 J

Work done W = 0 (for isochoric process)

Overall

?T = 243.7 - 182.7 = 61

q = q1 + q2

= 5.5 x 10^3J

w = w1 + w2 = -4.0 x 10^3J

?E = ?E1 + ?E2
= 1.5 x 10^3J
?H = ?H1 + ?H2

= 2.5 x 10^3J

From state A to stare B via lower path given in the question

Step 1

At constant volume (isochoric process)

Volume = 30 L

Pressure changes ?P = 1 atm - 2atm = - 1 atm

Change in temperature

?T = ?PV/nR

= - 1atm x 10 L / (2 mol x 0.0821 L-atm/mol-K )

= - 60.9 K

q1 = n x C v x ? T =  ?E1

= (2 mol) x (3/2R) x (-60.9 K)

q1 =?E1 = - 1.5 x 10 ^3 J

?H1 = n x Cp x ?T

= (2 mol) x (5/2R) x (-60.9 K)

=  - 2.5 x 10^3 J

Work done W = 0 (for isochoric process)

Step 2

At constant pressure (isobaric process) = 1atm

Volume change = 30 L - 10 L = 20 L

Work done = - P(V2 - V1) = - 1atm x 20 L

= - 20 L-atm x 101.325 J/L-atm

W = - 2026.5 J

For isobaric process

W = - nR (T2 - T1)

Change in temperature (T2 - T1)

= 2026.5 J / (2 mol x 8.314 J/mol·K)

= 121.8 K

At constant pressure

q 2 = n x Cp x ? T = ?H2

= (2mol) x (5/2R) x (121.8 K)

q2 = ?H2 = 5.1 x 10^3 J

? E2 = n x Cv x ? T

= (2mol) x (3/2R) x (121.8 K)

= 3.1 x 10^3 J

?T = -60.9+121.8

=60.9K
q = q1 + q2

= 3.6 x 10^ J
w = w1 + w2

= -2.0 x 10^3 J
?E = ?E1 + ?E2

= 1.5 x 10^3 J
?H = ?H1 + ?H2

= 2.5 x 10^3 J

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