A meteorological balloon contains 250.0 l he at 22°c and 740.0mmhg if the volume of the balloon can vary according to external conditions what volume would it occupy at an altitude at which the temperature is -52°c and the pressure is 0.750 atm

243.15 l

Explanation:

Using the general gas law

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/P₂T₁ where P₁, V₁ and T₁ are the initial pressure, volume and temperature of the meteorological balloon and  P₂, V₂ and T₂ are the initial pressure, volume and temperature of the meteorological balloon.

P₁ = 740 mmHg V₁ = 250 l and T₁ = 22°C + 273 = 295 K,

P₂ =  0.750 atm = 0.750 × 760 mmHg = 570 mmHg (since 1 atm = 760 mmHg) and T₂ = -52°C + 273 = 221 K  

V₂ = P₁V₁T₂/P₂T₁ = 740 mmHg × 250 l × 221 K/570 mmHg × 295 K = 243.15 l

V₂ = 243.15 l

So the volume it would occupy at an altitude of temperature -52 °C where it pressure is 0.750 atm  is 243.15 l

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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explanation: