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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