R = 0.821 ( atm Calculate the pressure in a 212 L tank containing 584 moles of Argon gas at 25℃? (Answer = 67.3 atm)

67.3 atm

Explanation:

We'll begin by converting 25 ℃ to Kelvin temperature. This can be obtained as follow:

T(K) = T(℃) + 273

T(℃) = 25 ℃

T(K) = 25 + 273

T(K) = 298 K

Finally, we shall determine the pressure. This can be obtained by using the ideal gas equation as illustrated below:

Gas constant (R) = 0.821 atm.L/Kmol

Volume (V) = 212 L

Number of mole (n) = 584 moles

Temperature (T) = 298 K

Pressure (P) =?

PV = nRT

P × 212 = 584 × 0.0821 × 298

P × 212 = 14288.0272

Divide both side by 212

P = 14288.0272 / 212

P = 67.3 atm

Thus, the pressure is 67.3 atm

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: