Question

The volume of a monatomic ideal gas doubles in an adiabatic expansion.

Considering 115 moles of gas with an initial pressure of 350 kPa and an initial volume of 1.4 m3 . Find the pressure of the gas after it expands adiabatically to a volume of 2.8 m3 .

Pf= 110 kPa

Find the temperature of the gas after it expands adiabatically to a volume of 2.8 m3 .

Answer #1

Number of moles n = 115 moles

Ratio of specific heat of monoatomic gas = 5/3

initial pressure P= 350 kPa 350000 Pa

initial volume V = 1.4 m^{3}

Final pressure P ' = ?

Final volume V = 2.8 m^{3} .

In adiabatic process , PV^{} = P ' V
'^{}

From this P ' = P ( V / V ' )

= 350 kPa (1.4/2.8) ^{5/3}

= 350 kPa (0.3149)

= 110.24 kPa

(b). Initial tempearture T = PV / nR

= (350000)(1.4) /(115)(8.314)

= 512.49 K

In adiabatic process TV^{-1} =
TV^{-1}

From this final temperature T ' = T (V/V ') -1

= 512.49 (1.4/2.8) ^{2/3}

= 322.85 K

Pf= 110 kPa

Find the temperature of the gas after it expands adiabatically to a volume of 2.8 m3

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