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A sample of methane of mass 4.50 g occupies 12.7 L

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Answer #1

given data,

m = 4.5 g

V1 = 12.7 L = 12.7*10^-3 m^3

T = 310 k


a) In Isobaric process pressure remains cosnatnt.

here, V2 = 12.7 + 3.3 L

= 16 L

= 16*10^-3 m^3

P = 30 kPa = 3*10^4 pa

Workdone in Isobaric process,
W = P*(V2 - V1)

= 3*10^4*3.3*10^-3

= 99 J

b) In Isothemrla process Temperature remains constant.

no of moles, n = mass/molar mass

= 4.5/16

= 0.28125 mole

Workdone in Isothermal process,
W = n*R*T*ln(V2/V1)

= 0.28125*8.314*310*ln(16/12.7)

= 167.4 J

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