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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 carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with P0V0 =13RT0  , the correct match is,


 

Moderate
Solution

Temperature at 1 : T03

Temperature at 2 : 2T03

Temperature at 3 : 3T03

 Work done in  12=P0V0                              23=0 Total work in   123=P0V0=13RToΔQ in              12=15R22T03-T03=56RToΔQ in              123=15R22T03-T03+13R23T03-2T03=43RToΔU in                 123=13R23T03-T03=RT0 



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Two moles of a monoatomic ideal gas undergoes a process AB as shown in the figure. Then find the work done during the process in Joule.


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