Question

When heated to 150 ºC, CuSO4.5 H2O loses its water of hydration as gaseous H2O. A 2.50 g sample of the compound is placed in a sealed 4.00 L steel vessel containing dry air at 1.00 atmand 27 ºC and the vessel is then heated to 227 ºC. What are the final partial pressure of H2O and the final total pressure

Answer 1

Water pressure 0.5 atm

Total Pressure= 2.27 atm

Step-by-step explanation:

To answer this problem, one has to realize that there are two processes that increase the temperature of the sealed vessel.

First, the dry air in the sealed vessel will be heated which will cause its pressure to increase and it can be determined by the equation:

P₁ x T₂ = P₂ x T₁ ∴ P₂ = P₁ x T₂ / T₁

For the second process, we have an amount of n moles of water which will be released when the copper sulfate is heated. In this case, to determine the value of the the water gas we will use the gas law:

PV = nRT ∴ P = nRT/V

n will we calculated from the quantity of sample.

2.50 g CuSo₄ 5H₂O x 1 mol/ 249.69 g = 0.01 mol CuSo₄ 5H₂O

the amount water of hydration is

= 0.01 mol CuSo₄ 5H₂O * 5 mol H₂O / 1 mol CuSo₄ 5H₂O

= 0.05 mo H₂O

pressure of dry air at the final temperature,

P₂ = 1 atm x 500 K/ 300 K = 1.67 atm

Pressure of water :

P (H₂O) 0.05 mol x 0.08206 Latm/kmol x 500 K/ 4 L = 0.5 atm

∴ Total Pressure = 1.67 atm

H2O Pressure = 0.5 atm