pretty sure it's both are physical changes.
Answer:
P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)
Explanation:
Given:
P₁ = 90 atm P₂ = ?
V₁ = 18 Liters(L) L₂ = 12 Liters(L)
=> decrease volume => increase pressure
=> volume ratio that will increase 90 atm is (18L/12L)
T₁ = 272 Kelvin(K) T₂ = 274 Kelvin(K)
=> increase temperature => increase pressure
=> temperature ratio that will increase 90 atm is (274K/272K)
n₁ = moles = constant n₂ = n₁ = constant
P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)
By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.
Answer : The vapor pressure of bromine at 
is 0.1448 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of bromine at = ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane = 
= normal boiling point of bromine = 
= heat of vaporization = 30.91 kJ/mole = 30910 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of bromine at is 0.1448 atm.
