Question

The volume of an ideal gas is adiabatically reduced from 217 L to 65.6 L. The...

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?

Homework Answers

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

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