Question

A mass of 222g of Helium gas at an initial temperature of 54.43°C and at an initial absolute pressure of 4.45 atm undergoes an isothermal expansion until its volume increases by a factor of 2.25. (a) What is the final pressure? (Pa)

(b) How much work is done on the gas?
(c) How much heat does the gas absorb?
(d) What is the change in the total internal energy of the gas?

Answer 1

Step-by-step explanation:

Given that,

Mass of gas = 222 g

Temperature = 54.43°

Pressure = 4.45 atm

Final volume = 2.25 initial volume

For isothermal expansion

(a). We need to calculate the pressure

Using relation of pressure and volume

`P_(i)V_(i)=P_(f)V_(f)`

`P_(f)=(P_(1)V_(i))/(V_(f))`

Put the value into the formula

`P_(f)=(4.45*V_(i))/(2.25V_(i))`

`P_(f)=(4.45)/(2.25)`

`P_(f)=1.97\ atm`

`P_(f)=199610.3\ pa`

`P_(f)=1.99*10^(5)\ Pa`

The pressure is
`1.99*10^(5)\ Pa`

(b). We need to calculate the work done

1 mole of Hg is 200.59 gram

222 g of Hg is

`n =(200)/(200.59)`

`n =0.997`

Using formula of work done

`W=nRTln((V_(f))/(V_(i)))`

Put the value into the formula

`W=0.997*8.314*(54.43+273)ln2.25`

`W=2200.9\ J`

`W=2.2009\ kJ`

The work done is 2.20 kJ.

(c). We need to calculate the gas absorb

Heat absorbed by the gas is the work done

`Q=W`

`Q=2.20 kJ`

The absorb heat is 2.20 kJ.

(d). We need to calculate the change in the total internal energy of the gas

Change in internal energy in an isothermal process is zero.

So,
`U=0`

Hence, This is the required solution.