from math import pi
#Variable Declaration Part a
vi = 20.0 #Initial volume of ideal gas, L
vf = 85.0 #final volume of ideal gas, L
Pext = 2.5 #External Pressure against which work is done, bar
#Calculations
w = -Pext*1e5*(vf-vi)*1e-3
#Results
print 'Part a: Work done in expansion is %6.1f kJ'%(w/1000)
#Variable Declaration Part b
ri = 1.00 #Initial diameter of bubble, cm
rf = 3.25 #final diameter of bubble, cm
sigm = 71.99 #Surface tension, N/m
#Calculations
w = -2*sigm*4*pi*(rf**2-ri**2)*1e-4
#Results
print 'Part b: Work done in expansion of bubble is %4.2f J'%w
#Variable Declaration Part c
i = 3.20 #Current through heating coil, A
v = 14.5 #fVoltage applied across coil, volts
t = 30.0 #time for which current is applied,s
from math import pi
#Calculations
w = v*i*t
#Results
print 'Part c: Work done in paasing the cuurent through coil is %4.2f kJ'%(w/1000)
#Variable Declaration Part d
k = 100.0 #Constant in F = -kx, N/cm
dl = -0.15 #stretch , cm
from math import pi
#Calculations
w = -k*(dl**2-0)/2
#Results
print 'Part d: Work done stretching th fiber is %4.2f J'%w
#Variable Declaration
m = 100.0 #Mass of water, g
T = 100.0 #Temperature of water, °C
Pext = 1.0 #External Pressure on assembly, bar
x = 10.0 #percent of water vaporised at 1 bar,-
i = 2.00 #current through heating coil, A
v = 12.0 #Voltage applied, v
t = 1.0e3 #time for which current applied, s
rhol = 997 #Density of liquid, kg/m3
rhog = 0.59 #Density of vapor, kg/m3
#Calculations
q = i*v*t
vi = m/(rhol*100)*1e-3
vf = m*(100-x)*1e-3/(rhol*100) + m*x*1e-3/(rhog*100)
w = -Pext*(vf-vi)*1e5
#Results
print 'Heat added to the water %4.2f kJ'%(q/1000)
print 'Work done in vaporizing liquid is %4.2f J'%w
#Variable Declaration Part d
m = 1.5 #mass of water in surrounding, kg
dT = 14.2 #Change in temperature of water, °C or K
cp = 4.18 #Specific heat of water at constant pressure, J/(g.K)
#Calculations
qp = m*cp*dT
#Results
print 'Heat removed by water at constant pressure %4.2f kJ'%qp
from math import log
#Variable declaration
n = 2.0 #moles of ideal gas
R = 8.314 #Ideal gas constant, bar.L/(mol.K)
#For reverssible Isothermal expansion
Pi1 = 25.0 #Initial Pressure of ideal gas, bar
Vi1 = 4.50 #Initial volume of ideal gas, L
Pf1 = 4.50 #Fianl Pressure of ideal gas, bar
Pext = 4.50 #External pressure, bar
Pint = 11.0 #Intermediate pressure, bar
#Calcualtions reverssible Isothermal expansion
T1 = Pi1*Vi1/(n*R)
Vf1 = n*R*T1/Pf1
w = -n*R*T1*log(Vf1/Vi1)
#Results
print 'For reverssible Isothermal expansion'
print 'Work done = %4.2e J'%w
#Calcualtions Single step irreverssible expansion
w = -Pext*1e5*(Vf1-Vi1)*1e-3
#Results
print 'For Single step reverssible expansion'
print 'Work done = %4.2e J'%w
#Calcualtions Two step irreverssible expansion
Vint = n*R*T1/(Pint)
w = -Pint*1e5*(Vint-Vi1)*1e-3 - Pf1*1e5*(Vf1-Vint)*1e-3
#Results
print 'For Two step reverssible expansion'
print 'Work done = %4.2e J'%w
from math import log
#Variable declaration
n = 2.5 #moles of ideal gas
R = 0.08314 #Ideal gas constant, bar.L/(mol.K)
cvm = 20.79 #Heat Capacity at constant volume, J/(mol.K)
p1 = 16.6 #Pressure at point 1, bar
v1 = 1.00 #Volume at point 1, L
p2 = 16.6 #Pressure at point 2, bar
v2 = 25.0 #Volume at point 2, L
v3 = 25.0 #Volume at point 3, L
#Calculations
T1 = p1*v1/(n*R)
T2 = p2*v2/(n*R)
T3 = T1 #from problem statement
#for path 1-2
DU12 = n*cvm*(T2-T1)
w12 = -p1*1e5*(v2-v1)*1e-3
q12 = DU12 - w12
DH12 = DU12 + n*R*(T2-T1)*1e2
#for path 2-3
w23 = 0.0
DU23 = q23 = n*cvm*(T3-T2)
DH23 = -DH12
#for path 3-1
DU31 = 0.0 #Isothemal process
DH31 = 0.0
w31 = -n*R*1e2*T1*log(v1/v3)
q31 = -w31
DU = DU12+DU23+DU31
w = w12+w23+w31
q = q12+q23+q31
DH = DH12+DH23+DH31
#Results
print 'For Path q w DU DH '
print '1-2 %7.2f %7.2f %7.2f %7.2f'%(q12,w12,DU12,DH12)
print '2-3 %7.2f %7.2f %7.2f %7.2f'%(q23,w23,DU23,DH23)
print '3-1 %7.2f %7.2f %7.2f %7.2f'%(q31,w31,DU31,DH31)
print 'Overall %7.2f %7.2f %7.2f %7.2f'%(q,w,DU,DH)
print 'all values are in J'
#Variable Declaration Part d
n = 2.5 #moles of ideal gas
R = 8.314 #Ideal gas constant, J/(mol.K)
cvm = 12.47 #Heat Capacity at constant volume, J/(mol.K)
pext = 1.00 #External Pressure, bar
Ti = 325. #Initial Temeprature, K
pi = 2.50 #Initial Pressure, bar
pf = 1.25 #Final pressure, bar
#Calculations Adiabatic process q = 0; DU = w
q = 0.0
Tf = Ti*(cvm + R*pext/pi)/(cvm + R*pext/pf )
DU = w = n*cvm*(Tf-Ti)
DH = DU + n*R*(Tf-Ti)
#Results
print 'The final temperature at end of adiabatic procees is %4.1f K'%Tf
print 'The enthalpy change of adiabatic procees is %4.1f J'%DH
print 'The Internal energy change of adiabatic procees is %4.1f J'%DU
print 'The work done in expansion of adiabatic procees is %4.1f J'%w
from math import log, exp
#Variable Declaration Part d
h1 = 1000.0 #initial Altitude of cloud, m
hf = 3500.0 #Final Altitude of cloud, m
p1 = 0.802 #Pressure at h1, atm
pf = 0.602 #Pressure at hf, atm
T1 = 288.0 #Initial temperature of cloud, K
cp = 28.86 #Specific heat of air, J/mol.K
R = 8.314 #Gas constant, J/mol.K
#Calculations
Tf = exp(-(cp/(cp-R)-1)/(cp/(cp-R))*log(p1/pf))*T1
#Results
print 'Final temperature of cloud %4.1f K'%Tf
if Tf < 273:
print 'You can expect cloud'
else:
print 'You can not expect cloud'