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answer asap please

A sealed balloon is filled with air at atmospheric pressure (P = 1.01 X 105 Pa) at 30°C (typical conditions for the surface o
3 pts. - d) The average molar mass of air molecules is M = 28.97 g moll. What is the average kinetic energy of air molecules
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Answer #1

From gas laws

\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

\frac{P_{1}T_{2}}{P_{2}T_{1}}=\frac{V_{2}}{V_{1}}

\frac{V_{2}}{V_{1}}=1.038

b)

we know PV=nRT

R = 0.082057 L-atm /mol K

putting P= 1 atm and T= 303 K

we get

n= 0.0201 moles

c)

now P = 0.95 atm

T =299 K

we get n= 0.0193

d)

We from from kinetic theory of gases

K.E= 3/2nRT

where K.E is kinetic energy

so K.E = 1.5 X 8.314 X 303

= 3778.713 J/mol

e)

for process under constant volume

molar heat capacity = Cv

given the gas is diatomic so Cv =5/2 R

now Q= nCv\Delta T

Q=7690.45 J

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