Question

1 mole of a gas undergoes a mechanically reversible isothermal expansion from an initial volume 1 liter to a final volume 10 liter at 25oC. In the process, 2.3 kJ of heat is absorbed in the system from the surrounding. The gas follows the following formula: V=RTP+b where V is the molar specific volume, and Tand Pare temperature (abosolute) and gas pressure respectively. Given R= 8.314 J/(mol.K) and b= 0.0005 m3. Evaluate the following a) Work (include sign) b) Change in internal energy of the gas c) Change in enthalpy of the gas In this situation wouldnt the change in enthalpy be H=Q+W=2300.1 please be clear with the work and description

Answer #1

given n = 1 mol

V1 = 1 l

V2 = 10 l

isothermal expansion, T = 25 C = 298 K

Q = 2.3 kJ

V = RTP + b

V is volume

P is pressure

T is temperature

R is gas constant

R = 8.314

b = 0.0005 m^3

a. work done in this expansion = W

now, dW = PdV

V = RTP + b

for isothermal process

dV = RTdP

dW = P*(RTdP)

integrating from P1 to P2

W = RT(P2^2 - P1^2)/2

W = ((V2 - b)^2 - (V1 - b)^2)/2RT

W = 1.81716*10^-8 J

b. change in internal energy = 0 J ( as the process is isothermal)

c. change in enthalapy = Q + w = 2300 + 1.81716*10^-8 = 2300.00000001 J

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