Question

A monatomic ideal gas with volume 0.230 L is rapidly compressed, so the process can be considered adiabatic. If the gas is initially at 1.01 105 Pa and 3.00 102 K and the final temperature is 489 K, find the work done by the gas on the environment, Wenv.

Answer 1

The value is
`W = - 17.53 \ J`

Step-by-step explanation:

From the question we are told that

The volume is
`V = 0.230 \ L = 0.230 *10^(-3) \ m^(-3)`

The initial pressure is
`P_1 = 1.01105 \ Pa`

The initial temperature is
`T_1 = 3.00*10^2 \ K`

The final temperature is
`T_2 = 489 \ K`

Generally for an adiabatic process the workdone is mathematically represented as

`W = - \Delta U`

Here
`\Delta U` is the internal energy of the system which is mathematically represented as

`\Delta U = (3)/(2) * nR \Delta T`

So

`W = - (3)/(2) * nR \Delta T`

Generally from ideal gas equation we have that

`n = (P_1V )/( RT_1 )`

Here R is the gas constant with value
`R = 8.314 J/mol\cdot K`

So

`n = (1.01 *0^(5) * 0.230 *10^(-3))/( 8.314 * 3.0*10^2 )`

=>
`n = 0.009313 \ mol`

So

`W = - (3)/(2) * 0.009313 * 8.314 * (451 - 3.00*10^2)`

=>
`W = - 17.53 \ J`