Question

The volume of an ideal gas is adiabatically reduced from 217 L
to 65.6 L. The initial pressure and temperature are 1.70 atm and
300 K. The final pressure is 9.07 atm. **(a)** Is the
gas monatomic, diatomic, or polyatomic? **(b)** What
is the final temperature? **(c)** How many moles are
in the gas?

Answer #1

Part A.

In adiabatic process we know that

PV^y = Constant

P1*V1^y = P2*V2^y

P1/P2 = (V2/V1)^y

take log both sides

ln (P1/P2) = y*ln (V2/V1)

y = ln (P1/P2)/ln (V2/V1)

Using given values:

y = ln (1.70/9.07)/ln (65.6/217)

y = 1.4 = 7/5 = Cp/Cv

So gas will be diatomic.

Part B.

In adiabatic process

T*V^(y - 1) = Constant

y - 1 = 1.4 - 1 = 0.4

T1*V1^0.4 = T2*V2^0.4

T2 = T1*(V1/V2)^0.4

T2 = 300*(217/65.6)^0.4

T2 = 484.11 K

Part C.

Using ideal gas law:

PV= nRT

n = PV/RT

P = 1.70 atm = 1.7*1.01325*10^5 Pa

V = 217 L = 0.217 m^3

T = 300 K

R = 8.314

So,

n = 1.70*1.01325*10^5*0.217/(8.314*300)

n = 14.98 = 15 moles

Please Upvote.

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