Question

One mole of an ideal gas at 300 K in thermal contact with the surroundings expands isothermally from1 L to2L against a constant pressure of 3 atm. in this process the change in the enthalapy of surroundings (change in S in J /K )1Latm.=101.3J

Answer 1

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Refer the following attachment.

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Answer 2

Explanation:

According to the first law of thermodynamics, energy cannot be created or destroyed as it can only be transformed from one form to another.

Also, relation between heat of the system, pressure and volume is as follows.

                 $q_{system}$ = $-P_{external} \times \Delta V$

As it is given that $P_{external}$ is 3 atm and change in volume ($\Delta V$) is 1 L. Putting these given values into the above formula as follows.

                 $q_{system} = -P_{external} \times \Delta V$

                                           =  $-3 atm \times (2 L -1 L)$

                                           = -3 L atm

The negative sign indicates that heat is releasing.

Also, relation between entropy and heat is as follows.

              $\Delta S_{system} = \frac{q_{rev}_{system}}{T}$

                                     = $\frac{3 \times 101.3 J}{300 K}$

                                     = 1.013 J/K

Hence, change in enthalpy of surrounding = $\Delta S_{surrounding}$ = $-\Delta S_{system}$ = -1.013 K.