Answer:
300.9mL
Explanation:
Given parameters:
V₁ = 280mL
T₁ = 22°C
T₂ = 44°C
Unknown:
V₂ = ?
Solution:
To solve this problem, we apply Charles's law;
it is mathematically expressed as;
We need to convert the temperature to kelvin;
T₁ = 22°C = 22 + 273 = 295K
T₂ = 44°C = 44 + 273 = 317K
Input the parameters and solve;
= 
V₂ x 295 = 280 x 317
V₂ = 300.9mL
Answer:
the change in energy of the gas mixture during the reaction is 227Kj
Explanation:
THIS IS THE COMPLETE QUESTION BELOW
Measurements show that the enthalpy of a mixture of gaseous reactants increases by 319kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -92kJ of work is done on the mixture during the reaction. Calculate the change of energy of the gas mixture during the reaction in kJ.
From thermodynamics
ΔE= q + w
Where w= workdone on the system or by the system
q= heat added or remove
ΔE= change in the internal energy
q=+ 319kJ ( absorbed heat is + ve
w= -92kJ
If we substitute the given values,
ΔE= 319 + (-92)= 227 Kj
With the increase in enthalpy and there is absorbed heat, hence the reaction is an endothermic reaction.
