The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are $6.21 \times 10^{-21}$ J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)
- 12.42 $\times 10^{-21}$ J, 968 m/s
- 8.78 $\times 10^{-21}$ J, 684 m/s
- 6.21 $\times 10^{-21}$ J, 968 m/s
- 12.42 $\times 10^{-21}$ J, 684 m/s
The Correct Option is D
Solution and Explanation
$ \, \, \, \, \, \, \, \, \, v_{rms}=\sqrt{\frac{3RT}{M}} \, i.e.,v_{rms} \, = \sqrt T$
When temperature of gas is increased from 300 K to 600 K
(i.e. 2 times), the average translational KE will increase to \2 times and rms speed to $\sqrt 2 $ or 1.414 times.
$\therefore \, \, \, $ Average translational KE =$2 \times 6.21 \, \times 10^{-21} J$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =12.42 \times 10^{-21} J$
and $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, v_{rms}=(1.414)(484)m/s $
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =684 m/s $
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Concepts Used:
Spontaneity
Spontaneity comes under the First Law of Thermodynamics that is based on the law of conservation of energy which explains that energy can be transformed from one form to another but cannot be created or destroyed. The spontaneity in thermodynamics defines the direction of heat flow that can be developed by establishing a relationship between the work done by the system or by the system. All of the processes of heat flow which happen naturally tend to proceed spontaneously only in one direction.
Spontaneous Reaction
A spontaneous chemical reaction is an irreversible process where you can’t get the ingredients back without the external agents.
Total entropy change is the essential parameter that defines the spontaneity of any process. Since most of the chemical reactions fall under the category of a closed system and open system; we can say there is a change in enthalpy too along with the change in entropy. Since, change in enthalpy also increases or decreases the randomness by affecting the molecular motions, entropy change alone cannot account for the spontaneity of such a process. Therefore, for explaining the spontaneity of a process we use the Gibbs energy change. Gibbs’ energy is a state function and an extensive property. The general expression for Gibbs energy change at constant temperature is expressed as:
ΔGsys = ΔHsys – TΔSsys
Here,
Change in Gibbs energy of the system = ΔGsys
Change in enthalpy of the system = ΔHsys
Change in Entropy of the system = ΔSsys
Constant Temperature of the system = T
Also, if we conduct a spontaneous process, the total change in entropy is always greater than zero.
Mathematical expression for the above spontaneous reaction meaning expression is
ΔSsys + ΔSsurr = ΔStotal
Here,
ΔStotal = total change in entropy for the process
ΔSsurr = change in entropy of the surrounding
ΔSsys = change in entropy of the system
Also, for a spontaneous process, the total change in entropy is 0, i.e. ΔStotal> 0.
Therefore;
TΔSsys – ΔHsys>0
ΔHsys– TΔSsys<0
Using the Gibbs equation, it can be said that
ΔGsys< 0
Thus, it can be inferred that any process is spontaneous if the change in Gibbs energy of the system is less than zero or else the process is not spontaneous.
This by the already provided equations the spontaneity can be predicted.
- During an exothermic reaction, the system has a negative enthalpy which inturn makes all the exothermic reactions spontaneous.
- During an endothermic reaction when the temperature or the entropy change is extremely high then the Gibbs free energy becomes negative.