I think its A because some scientist from the Department of Physics of Northeastern University found out that is not a part of Dalton's atomic theory.
Answer: they giving you some hard question
Explanation:
i dont know know what the answer is do you have any answer options
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:

We know that,
The relation between the
for an ideal gas are :

As we are given :



Now we have to calculate the entropy change of the gas.


(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Answer:
Never smoke near a place where hydrogen is generated or being used. Remove all possible sources of flame and sparks. Hydrogen should only be generated and used in a well ventilated out door area. Precautions must be taken to remove all possibilities of fire or explosion.