Question

two moles of a monatomic ideal gas are compressed in a cylinder at a constant temperature of 85 c until the original pressure has tripled?

a)what is the work done on the gas?

b)How much heat is transfered out of the gas?

A monatomic ideal gas in a cylinder is held at a constant temperature 230kpa and is cooled and compressed from 1.7 to 1.2

a) what is the internal energy of the gas?

b)How much heat is transferred out of the gas?

Answer 1

1) a) Since, the temperature is a constant the process is isothermal

Using ideal gas equation.

For isothermal process

Substituting values n= 2 ;R= 8.314 j/mol. k ; T=85+273= 358 k

b) Acoording to First law of thermodynamics.

Since (no change in temp.)

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2)a) Based on the update : Assuming constatn pressure and Volume changed from 1.7 to 1.2 liters during compression.

For a monoatomic ideal gas the only contribution to the internal energy comes form the translational kinetic energy.

Using Ideal gas law ;

PV=nRT

T=PV/(nR)

Substituting For T from ideal gas equation

(Internal energy decreases due to cooling process)

b) Work done = PdV=-115 J

Using 1st law Q = -172.5-115=- 287.5 J (Heat flows out of the gas)

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