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

Answer:

The value is  $W = - 17.53 \ J$    

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

                $\Delta U = \frac{3}{2} * nR \Delta T$

So      

               $W = - \frac{3}{2} * nR \Delta T$

Generally from ideal gas equation we have that

               $n = \frac{P_1V }{ RT_1 }$

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

So

                 $n = \frac{1.01 *0^{5} * 0.230 *10^{-3}}{ 8.314 * 3.0*10^2 }$

=>              $n = 0.009313 \ mol$

So  

                 $W = - \frac{3}{2} * 0.009313 * 8.314 * (451 - 3.00*10^2)$    

=>             $W = - 17.53 \ J$