Question

A gas in a piston–cylinder assembly undergoes a compression process for which the relation between pressure and volume is given by pVn = constant. The initial volume is 0.1 m3 , the final volume is 0.04 m3 , and the final pressure is 2 bar. Determine the initial pressure, in bar, and the work for the process, in kJ, if (a) n = 0, (b) n = 1, (c) n = 1.3

Answer 1

Answer:

A) P1=2 [bar] , W=-12 [kJ]

B) P1=0.8 [bar] , W=-7.3303 [kJ]

C) P1=0.6077 [bar] , W=-6.4091 [kJ]

Explanation:

First, from the problem we know the following information:

V1=0.1 m^3

V2=0.04 m^3

P2=2 bar =200 kPa

The relation PV^n=constant means PV^n is a constant through all the process, so we can derive the initial pressure as:

$P_{1}V^{n} _{1}= P_{2}V^{n} _{2}$

$P_{1}= \frac{P_{2}V^{n} _{2}}{V^{n} _{1}}$

a) To the case a) the constant n is equal to 0, we can calculate the initial pressure substituting n=0 in the previous expression, so:

$P_{1}= \frac{P_{2}V^{n} _{2}}{V^{n} _{1}}=\frac{(200 kPa)(0.04 m^{3})^{0} }{(0.1 m^{3})^{0} }=200 kPa=2 bar$

The expression to calculate the work is:

$W=\int\limits^2_1 {P} \, dV$

If n=0:

$P_{1}V_{1}^0 =P_{2}V_{2}^0 \\P_{1}=P_{2}$

Then:

$W=\int\limits^2_1 {P} \, dV=P\int\limits^2_1 {} \, dV=P(V_{2}-V_{1})$

The work is:

$W=P(V_{2}-V_{1})=(200 kPa)(0.04 m^3 -0.1 m^3)=-12 (kPa)(m^3)=-12kJ$

b) To the case b) the constant n is equal to 1, we can calculate the initial pressure substituting n=1 in the initial expression, so:

$P_{1}= \frac{P_{2}V^{n} _{2}}{V^{n} _{1}}=\frac{(200 kPa)(0.04 m^{3})^{1} }{(0.1 m^{3})^{1} }=80 kPa=0.8 bar$

If n=1 then:

$P_{1}V_{1}^1 =P_{2}V_{2}^1 \\P_{1}V_{1} =P_{2}V_{2}=constant$

To calculate the work:

$[tex]W=constant\int\limits^2_1 {\frac{1}{V} } \, dV$[/tex]

$W=(constant)ln(\frac{V_{2}}{V_{1} } )=PVln(\frac{V_{2}}{V_{1} } )$

Substituting:

$W=(80kPa)(0.1 m^3)ln(\frac{0.04 m^3}{0.1m^3 } )=-7.3303kJ$

c) To the case c) the constant n is equal to 1.3, we can calculate the initial pressure substituting n=1.3 in the initial expression, so:

$P_{1}= \frac{P_{2}V^{n} _{2}}{V^{n} _{1}}=\frac{(200 kPa)(0.04 m^{3})^{1.3} }{(0.1 m^{3})^{1.3} }=60.7726 kPa=0.6077 bar$

First:

$PV^n =constant\\P=constant V^{-n}$

The work:

$W=\int\limits^2_1 {P} \, dV=\int\limits^2_1 {constant V^{-n}} \, dV\\W=(constant)(\frac{V^{-n+1}}{-n+1} )\left \{ {{2} \atop {1}} \right. \\W=(constant)(\frac{V_{2}^{-n+1}-V_{1}^{-n+1}}{-n+1} )\\W=(PV^n)(\frac{V_{2}^{-n+1}-V_{1}^{-n+1}}{-n+1} )\\W=\frac{(P_2V_2-P_1V_1)}{1-n}$

Substituting:

$W=\frac{(P_2V_2-P_1V_1)}{1-n}=\frac{(200kPa)(0.04m^3)-(60.7726kPa)(0.1m^3)}{1-1.3}$

W=-6.4091 kJ