Question

Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected to an evacuated 20.0 L sealed bulb via a valve. Assume that the temperature remains constant.a.What should happen to the gas when you open the valve?b.Calculate ΔH, ΔE, q, and w for the process you described above.

Answer 1

Complete Question

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Answer

a

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

b

$\Delta H = 0\ , \Delta E = 0 , q= 0 , w= 0$

c

The driving force for this process is the increase in entropy this is because the movement of the internal energy of the gas into a larger volume, what this does is that it increases the amount of disorder(entropy).

Explanation

In order to obtain the parameter in the part B of the question we are first obtain the initial pressure, using the ideal gas equation  

                      $P = \frac{nRT}{V}$

                     $P = \frac{(2.4mol)(0.0821\frac{1 atm}{K \cdot \ mol} )}{4.0L}$

                         $P =15 \ atm$

The next thing is to obtain the new pressure of the gas , using boyle's law

              $P_1V_1 = P_2V_2$

                  $P_2 = \frac{P_1 V_1}{V_2}$

                  $P_2 = \frac{(15 \ atm)(4.0L)}{24.0 L}$

                  $P_2 = 2.5 \ atm$    

Since the this process is isothermal , the change in heat is equal to zero

                      i.e  q = 0 J

  The workdone to move  the gas to the other container is zero because the  the pressure at this second container is zero due to the fact that it is a vacuum

    i.e  $w = -P_{external} \Delta V$

              $=-(0 \ atm) (24.0 - 4.0L)$

              $= 0L \cdot atm$

  Since the change in heat is zero and the workdone is zero then the change in internal energy is equal to 0

     This is because the change in internal energy is equal to a summation of change in heat and the workdone

                i.e $\Delta E = q + w$

                            $= 0J$

Generally the change in enthalpy is mathematically represented as

                $\Delta H = n C_p \Delta T$

Since the temperature is zero this means that the change in temperature is zero , substituting this value for change in temperature into the equation for  change in enthalpy

           $\Delta H = n C_p (0)$

                   $= 0J$