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## A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 at

Question

A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 atm.(a) Determine the initial and final temperatures.initial Kfinal K(b) Determine the change in internal energy. J(c) Determine the heat lost by the gas. J(d) Determine the work done on the gas. J

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2021-07-15T12:28:45+00:00
2021-07-15T12:28:45+00:00 1 Answers
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## Answers ( )

Answer:a)Initial Temperature =609.4 Kand Final Temperature =325.7 Kb)the change in internal energy is-18279.78 Jc)heat lost by the gas iszeroor0d)the work done on the gas is-18279.78 JExplanation:Given the data in the question;

P = 1 atm = 101325 pascal

P = ?

V = 0.1550 m³

V = 0.742 m³

we know that for an adiabatic process γ = 1.4

PV = PV

P = P[tex]([/tex] V / V

we substitute

P = 1 × 0.1550 / 0.742

= 0.2088948787

= 0.11166 atm

a) the initial and final temperatures

Initial temperature

T = PV / nR

given that n = 3.10 mol

= ( 101325 × 0.1550 ) / ( 3.10 × 8.314 )

= 15705.375 / 25.7734

T =

609.4 KFinal temperature

T = PV / nR

= ( 0.11166 × 101325 × 0.742 ) / ( 3.10 × 8.314 )

= 8394.95 / 25.7734

=

325.7 KTherefore, Initial Temperature =

609.4 Kand Final Temperature =325.7 Kb) the change in internal energy

ΔE = nCΔT

here, C = ( 5/2 )R

ΔE = 3.10 × ( 5/2 )8.314 × ( 325.7 – 609.4 )

=

-18279.78 JTherefore, the change in internal energy is

-18279.78 Jc) the heat lost by the gas

Since its an adiabatic process,

Q =

0Therefore, heat lost by the gas is

zeroor0d) the work done on the gas

W = ΔE – Q

= -18279.78 J – 0

W =

-18279.78 JTherefore, the work done on the gas is

-18279.78 J