Question

Assume air can be modeled as an ideal, diatomic gas. Given n moles of air
at temperature T0 and pressure p0, how much energy must be expended as work to
compress it to one tenth of its original volume, if
a) the compression occurs along an isotherm?
b) the compression occurs along a reversible adiabatic path?
c) Evaluate your results to parts (a) and (b) for n = 5.00 mol, T0 = 20.0
◦C, and
p0 = 1.00 atm. Express your answers in kilojoules (kJ)

Answer 1

let the initial volume of the gas = $v_{1}$ = v

fibal volume = $v_{2}$ = v / 10

since the gas is diatomic ,n=2

temperature = $T_{o}$

pressure = $P_{o}$

part-a:

work done im isothermal compression of gas is given by

$w=nRT\ln (\frac{v_{2}}{v_{1}})$

=> $w=nRT\ln (\frac{\frac{v}{10}}{v})$

=>    $w=nRT\ln (0.1)$

=>    $w=-2.302nRT_{o}$ ..... equation-2

=> $w=-4.604RT_{o}$

part-b:

for adiabatic process $Pv^{\gamma }=k$

by subtituting initial conditions    $k=P_{0}v^{\gamma }$

we know Pv=nRT

by subtituting initial values,the initial volume of gas is given by

$v=nRT_{o}/P_{o}$    ...equation-1

for reversible adiabaticprocees work done is given by

$w=\frac{k(v_{2}^{1-\gamma }-v_{1}^{1-\gamma })}{1-\gamma }$

=> $w=\frac{P_{o}v^{\gamma }(v_{2}^{1-\gamma }-v_{1}^{1-\gamma })}{1-\gamma }$

=> $w=\frac{P_{o}v^{\gamma }((\frac{v}{10})^{1-\gamma }-v^{1-\gamma })}{1-\gamma }$

=>    $w=\frac{P_{o}v(0.1^{\gamma }-1)}{1-\gamma }$

=>    $w=\frac{nRT_{o}(0.1^{\gamma }-1)}{1-\gamma }$   ..... equation 2

for diatomic gas $\gamma =\frac{7}{5}$

=>    $w=4.8RT_{o}$

part-c:

$n=5$ ,   ,   ,

from equation-1

isothermal work = -2.302(5)(0.082)(293) joules

=276.54 joules

=0.27 kjoules

from equation-2

for penta atomic gases $\gamma =1.31$

reverse adiabatic work = $5(0.082)(293)(\frac{0.1^{1.31}-1}{1-1.31})$

=368.54 joules

=0.37 kjoules