Question

The molar heat capacity CP,m of SO2(g) is described by the following equation over the range 300 K < T < 1700 K :
CP,mR = 3.093+6.967×10−3TK
−45.81×10−7T2K2+1.035×10−9T3K3
In this equation, T is the absolute temperature in kelvin. The ratios Tn/Kn ensure that CP,m has the correct dimension. 1.40 moles of SO2(g) is heated from 24.0 ∘C to 1.11×10³ ℃ at a constant pressure of 1 bar.

Assuming ideal gas behavior, calculate ΔH

Assuming ideal gas behavior, calculate q.

Assuming ideal gas behavior, calculate ΔU

Assuming ideal gas behavior, calculate w.

Answer 1

a)

dH = n*Cp*dT

Cp(funciton of T) = 3.093+(6.967*10^3)*T1 −(45.81*10^7)*T2 +(1.035*10^9)*T3

we must interate

Hf-Hi = 1.4 mol *(3.093*T+(6.967*10^3)/2*T1^2 −(45.81*10^7)/3*T2^3 +(1.035*10^9)/4*T3^4) from Ti to Tf

evaluate Ti and Tf

dH = (1.4mol)(3.093*Tf+(6.967*10^3)/2*(Tf^2) −(45.81*10^7)/3*(Tf^3) +(1.035*10^9)/4*(Tf^4) - ((3.093*Ti+(6.967*10^3)/2*(Ti^2) −(45.81*10^7)/3*(Ti^3) +(1.035*10^9)/4*(Ti^4) )

substitute

T1 = 24+273 = 297 K

T2 = (1.11*10^3) + 273 = 1383 K

dH = (1.4mol)(3.093*1383 +(6.967*10^3)/2*(1383 ^2) −(45.81*10^7)/3*(1383 ^3) +(1.035*10^9)/4*(1383 ^4) - ((3.093*297 +(6.967*10^3)/2*(297 ^2) −(45.81*10^7)/3*(297 ^3) +(1.035*10^9)/4*(297 ^4) )

dH = 1.4*( (9.4620*10^20) - (1.9556*10^-18))

dH = 1.324*10^21 J

b)

for Q

since this is a closed system at constant P

then

Q - W = dH

no work so

Q = dH

Q = 1.324*10^21 J

d)

W = -P*dV

W = - 100000 Pa * (Vf-Vi)

PV = nRT

V = nRT/P = (1.4)(0.082)(297 )/(0.988) = 34.27 L

since n, and P are constant

P1V1/(n1T1) = P2V2/(n2T2)

converts to

V1/T1 = V2/T2

T1 = 24+273 = 297 K

T2 = (1.11*10^3) + 273 = 1383 K

V1/T1 = V2/T2

V2 = V1*T2/T1 = 34.27 L * (1383 /295 / ) = 160.66240 L

change in V

dV = V2-V1 = 160.66240-34.27 = 126.3924 L = 126.3924*10^-3 m3

W = -(10000 Pa)(126.3924*10^-3 m3) = 126.3924 J

W = 126.3924 J

c)

dH = PdV + dU

the change in volume:

dU = dH - P*dV

dU = dH - P*dV

dU = 1.324*10^21 J - 126.3924 J = 1.324*10^21 J