First slide
Heat ;work and internal energy
Question

Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by TΔX , where T is temperature of the system and ΔX is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas X=32RlnTTA+RlnVVA . Here R is gas constant, V is volume of gas and TA and VA are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

List-IList-II
I) Work done by the system in process 123P) 13RT0ln2
II) Change in internal energy in process  123Q) 13RT0
III) Heat absorbed by the system in process  123R)  RT0
IV) Heat absorbed by the system in process  12S) 43RT0
 T) 13RT03+ln2
 U) 56RT0

 

If the process on one mole of monatomic ideal gas is as shown in the TV-diagram with P0V013RT0 , the correct match is,


 

Moderate
Solution

 Work done in     12=1RT03ln2V0V0=RTo3ln2                               23=0Total work done =RTo3ln2ΔU in                  12=0ΔU in                  23=13R2T0-T03=RToTotal change in internal energy=RToΔQ in                  12=ΔW+ΔU=RTo3ln2ΔQ in                  23=ΔW+ΔU=RToTotal heat absorbed=RTo+RTo3ln2=13RT03+ln2 

Get Instant Solutions
When in doubt download our app. Now available Google Play Store- Doubts App
Download Now
Doubts App