Question

A 0.450-mol sample of an ideal diatomic gas at 372 kPa and 312 K expands quasi-statically until the pressure decreases to 147 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is the following. 

(a) Isothermal 

final temperature _______  

volume of the gas _______  

work done by the gas _______  

heat absorbed _______  

(b) adiabatic 

final temperature _______  

volume of the gas _______  

work done by the gas _______  

heat absorbed _______ 

Answer 1

a) (Isothermal Expansion)

Since the expansion is isothermal, temperature remains the same. That is,

Final temperature :312K

The initial and final volumes and pressure in an isothermal process is related as,

Given that,

The initial volume of the gas is found by using the ideal gas equation,

Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature.

Then,

That is,

The final volume is,

Volume of the gas is : 7.941L

Work done by the gas is given by,

That is,

Work done by the gas : 1083.82 J

In an isothermal process, the heat absorbed is equal to the work done. Therefore,

Heat absorbed : 1083.82 J

b) (Adiabatic Expansion)

In adiabatic expansion, .

Then, we can write,

Where,

Then,

That is,

Final temperature=239.30K

In an adiabatic process, we have

Or we can write,

We have,

Then,

Volume of the gas : 6.091L

The work done is given by,

On substituting the values, we get,

Work done by the gas : -680.006 J

Since it is an adiabatic process, there is no heat transfer.

That means,

Heat absorbed : Zero