Chemistry Class-11: Chapter –6. Thermodynamics Part – 7
Glide to success with Doorsteptutor material for CBSE : fully solved questions with step-by-step explanation- practice your way to success.
Question: 45
What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, in a single step as shown in Fig. 6.2. Explain graphically.
PV in Cylinder
Answer:
Work done on an ideal gas can be calculated from graph shown in Fig. 6.6. Work done is equal to the shaded area .
PV Curve Area
Question: 46
How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?
Answer:
The work done can be calculated with the help of plot of the work of compression which is carried out by change in pressure in infinite steps, is given in Fig. 6.7. Shaded area represents the work done on the gas.
PV Curve
Question: 47
Represent the potential energy/enthalpy change in the following processes graphically.
(a) Throwing a stone from the ground to roof.
(b)
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Answer:
Distance vs. Energy
Fig.: 6.8 Enthalpy change in processes (a) and (b)
Question: 48
Enthalpy diagram for a particular reaction is given in Fig. 6.3. Is it possible to decide spontaneity of a reaction from given diagram. Explain.
Answer:
No.
Enthalpy is one of the contributory factors in deciding spontaneity but it is not the only factor. One must look for contribution of another factor i.e., entropy also, for getting the correct result.