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Varvara68 [4.7K]

2 years ago

8

A balloon that can hold 85 L of air is inflated with 3.5 moles of gas at a pressure of 1.0 atmosphere. What is the temperature o

f the balloon?

1 answer:

slega [8]2 years ago

7 0

Answer:

295.96 K ( = 22.81 degrees C)

Explanation:

PV = n RT R = .082057 L-Atm / (K-Mol) n = 3.5 P = 1 atm V = 85 liters

(1)(85) = 3.5 (.082057)(T)

solve for T = 295.96 K

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The process involved in this problem are :

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$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

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$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

Now put all the given values in the above expression, we get:

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

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