Question

(a) How much work is required to compress 4.92 mol of air at 19.9

Answer 1

(a)
Generally the work done on the gas is given by the integral
W = - ? p dV from initial to final state
To solve it you need to relate pressure p to volume V.

For an isothermal process on an ideal gas, ideal gas law states that:
p?V = n?R?T = constant
=>
p = n?R?T/V

So the work required to change volume from V? to V? is
W = - ?V? to V? [ n?R?T/V ] dV
= - n?R?T? ?V? to V? [ 1/V ] dV
= - n?R?T?ln(V?/V?)
= n?R?T?ln(V?/V?)

For this process
W = 4.96mol ? 8.314472J/molK ? (19+273.15)K ? ln(12) = 29939.67J = 29.94kJ

(b)
For an adiabatic process on ideal gas [1]
p?V^? = constant

?=cp/cv is the heat capacity ratio. For air, which consist mainly of diatomic nitrogen and oxygen,
it takes the same value as for an diatomic gas.
? = 7/5 = 1.4

Using this relation you can express pressure in terms of initial pressure and volume:
p?V^? = constant = p??V?^?
=>
p? = p??(V?/V)^?

Hence,
W = - ?V? to V? [ p??(V?/V)^? ] dV
= - p??V?^? ? ?V? to V? [ V^(-?) ] dV
= - p??V?^? ? ( V?^(1-?) - V?^(1-?) ) / (1-?)
= p??V?^? ? ( V?^(1-?) - V?^(1-?) ) / (? - 1)
< factor out V?^(1-?)
= p??V? ? ( (V?/V?)^(1-?) - 1) ) / (? - 1)
< with p??V? = n?R?T?
= n?R?T? ? ( (V?/V?/)^(?-1) - 1) ) / (? - 1)

For our process
W = 4.96mol ? 8.314472J/molK ? (19+273.15)K ? ((12)^(0.4) - 1))/0.4 = 51262.69J = 51.26kJ


(c)
isothermal
p?V = constant
=>
p??V? = p??V?
=>
p? = p??(V?/V?) = 1.0atm ? 12 = 12atm

adiabatic
p?V^? = constant
=>
p??V?^? = p??V?^?
=>
p? = p??(V?/V?)^? = 1.0atm ? 12^1.4 = 32.42atm