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

The complete question is shown on the first uploaded image

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).

Step-by-step 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 = (nRT)/(V)`

`P = ((2.4mol)(0.0821(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 = (P_1 V_1)/(V_2)`

`P_2 = ((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`