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Solution


Q.

The temperature of a gas having 2.0×1025 molecules per cubic meter at 1.38 atm is

(Given, k=1.38×10-23JK-1)                    [2024]
1

300 K

 

2

500 K

 

3

100 K

 

4

200 K

 


Ans.

(2)   

        ldeal gas equation,

         PV=NNART

         N = Total no. of molecules

         P=NVkT

         1.38×1.01×105=2×1025×1.38×10-23×T

         ⇒T=1.01×1032≈500K