According to ideal gas equation, we know for 1 mole of gas: PV=RT
where P = pressure, T = temperature, R = gas constant, V= volume
If '1' and '2' indicates initial and final experimental conditions, we have

Given that: V1 = 100.0 kPa, T1 = 100.0 K, V1 = 2.0 m3, T2 = 400 K, P2 = 200.0 kPa
∴ on rearranging above eq., we get V2 =

∴ V2 = 4 m3
Answer:
the molarity is 3.68 moles/L
Explanation:
the molality of the solution of sucrose is
m= moles of glucose / Kg of solvent (water)= 6.81 ,
since the molecular weight of glucose is 180.156 gr/mole , then per each kilogram of solvent there is
6.81 moles*180.156 gr/mole + 1000 gr of water = 2226.86 gr of solution
from the density
volume of solution = mass of solution/density = 2286.86 gr / 1.2 gr/ml = 1855.71 ml
therefore there is 1000 gr of water in 1855.71 ml
then the molarity M is
M= moles of glucose / L of solution = (moles of glucose / Kg of solvent) * (Kg of solvent/L of solution) = 6.81 moles/Kg * 1Kg/1.85 L = 3.68 moles/L
M= 3.68 moles/L
Note:
- Would be wrong in this case to assume density of water = 1 Kg/L since the solution is heavily concentrated in glucose and therefore the density of water deviates from its pure value.
Answer:
I think the answer is increases