Discussion Forum

One mole of an ideal gas at 300K  in thermal contact with surroundings expands isothermally from 1.0 L  to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy  of surrounds ( $\triangle S_{surr}$ ) in J  $K^{-1}$  is (1L atm=101.3 J)

Answer:

Explanation:

By the first law.$\triangle  E=Q+W$

 For isothermal expansion, $\triangle  E=0$ 

  $\therefore$ Q=-W

    $-Q_{irrev}=W_{irrev}=p\triangle V$

                      =3(2-1)= 3L atm

   Also,  $\triangle S_{surr}= \frac{Q_{irrev}}{T}=\frac{(-3\times101.3)J}{300 K}$

                                    $= -\frac{303.9}{300}=-1.013 JK^{-1}$