In [3]:

```
# Variables
l = 80; #latent heat of fusion in cal
j = 4.2*10**7; #joules constant in ergs/cal
w = -0.092*10**6; #work done in changing phase change in ergs
# Calculations
q = l*j; #heat added in ergs
du = q-w; #internal energy in ergs
# Result
print 'the change in internal energy is %.1e ergs'%(du)
```

In [5]:

```
# Variables
m = 1.; #mass in gm
l = 536.; #latent heat in cal/gm
j = 4.2*10**7; #joules constant in ergs/cal
v = 1649; #volume of water in cc
p = 76*13.6*981; #pressure of water in dynes/sq.cm
# Calculations
dq = m*l*j; #heat supplied in ergs
dw = p*v; #work done in ergs
du = dq-dw; #internal energy developed in ergs
# Result
print 'internal energy of water is %.2E ergs'%(du)
```

In [7]:

```
# Variables
dv = 10; #ratio of original volume to final volume
t1 = 293; #inital temperature in K
y = 1.41; #coefficent of expansion
# Calculations
t2 = t1*(dv)**(y-1); #final temperature in K
# Result
print 'the final temperature is %.f K'%(t2)
```

In [8]:

```
# Variables
t = 273; #temperature of earth at height h in K
p = 760; #pressure in mm of hg
dp = 1; #change in pressure in mm of hg
y = 1.418; #coefficient of expansion
# Calculations
dt = ((y-1)/y)*dp*t/p; #change in temperature in deg.C
# Result
print 'the change in temperature is %3.3f deg.C'%(dt)
```

In [12]:

```
# Variables
p1 = 2; #pressure initial in atm
p2 = 1.; #pressure final in atm
t1 = 273 + 15; #inital temperature in K
y = 1.4; #coefficent of expansion
# Calculations
t2 = t1*(p2/p1)**((y-1)/y); #final temperature in K
dt = t1-t2; #drop in temperature in K
# Result
print 'drop in temperature is %3.2f K'%(dt)
print "Note : answer is slightly different because of rounding error"
```

In [13]:

```
# Variables
t1 = 288; #inital temperature in K
dv = 1./2; #ratio of inital to final volume
y = 1.4; #coefficient of expansion
# Calculations
t2 = t1*(dv)**(y-1); #final temperature in K
# Result
print 'the final temperature is %3.1f K'%(t2)
```

In [14]:

```
# Variables
y = 1.4; #coefficent of exapnsion
p1 = 1; #standard pressure in atm
dv = 50; #ratio of initial volume to final volume
t1 = 273; #standard temperature in K
# Calculations
p2 = p1*dv; #final pressure when slowly compressed in atm
p3 = p1*(dv)**(y); #final pressure when suddenly compressed in atm
t2 = t1*(dv)**(y-1); #rise in temperature when it is suddenly compressed in K
# Result
print 'the final pressure when it is compressed slowly is %.f atm \
\nthe final pressure when it is compressed suddenly is %.f atm \
\nthe rise in temperature when it is suddenly compressed is %.0f K'%(p2,p3,t2)
```

In [15]:

```
# Variables
y = 1.5; #coefficient of expansion
dp = 1./8; #ratio of inital pressure to final pressure
t1 = 300; #inital tempreature in K
# Calculations
t2 = t1*(dp)**((1-y)/y); #change in temperature in K
t3 = t2-t1; #rise in temperature in K
# Result
print 'the rise in temperature is %3.2f K'%(t3)
```

In [17]:

```
import math
# Variables
t1 = 400; #inital temperature in K
dv = 2; #ratio of volumes final and inital
r = 8.31*10**7; #universal gas constant in ergs/kg.K
# Calculations
w = r*t1*math.log(2); #work done in expanding isothermally in ergs
# Result
print 'the work done in expanding isothermally is %.1e ergs'%(w)
```

In [21]:

```
# Variables
p1 = 76; #inital pressure in cm
t1 = 290; #inital temperature in K
y = 1.4; #coefficent of expansion
dv = 2; #ratio of inital to fianl volume when air expands isothermally
dv1 = 2; #ratio of inital to final volume when air expands adiabatically
# Calculations
p2 = p1/dv; #final pressure when air expands isothermally in cm of hg
t2 = t1; #final temperature when air expands isothermally in K
t3 = t2*(1./dv1)**(y-1); #temprature when air expands adiabatically in K
p3 = p2*(1./dv1)**(y); #final pressure when air expands adiabatically in mm of hg
# Result
print 'final pressure when air expands isothermally in cm of hg %3.2f mm of hg \
\nfinal temperature when air expands isothermally is %3.2f K \
\ntemprature when air expands adiabatically is %3.1f K \
\nfinal pressure when air expands adiabatically is %3.2f cm of mercury'%(p2,t2,t3,p3)
```

In [25]:

```
# Variables
p = 76*13.6*981; #pressure of air in dynes/sq.cm
v = 11100.; #volume expanded in ml
t1 = 273.; #inital temperature in K
t2 = 274.; #final temperature in K
cv = 2.411; #specific heat at constant volume in cal/K
j = 4.2*10**7; #joules constant in ergs/cal
# Calculations
w = p*v*math.log(t2/t1); #work done in ergs
h = cv*(t2-t1)+w/j; #heat supplied in cal
# Result
print 'the work done is %3.3e erg \
\nthe heat supplied is %3.3f cal'%(w,h)
print "Note : answer is different because of rounding error"
```

In [31]:

```
# Variables
p = 10**6; #pressure of air in dynes
d = 0.0001293; #density of air in gm/cc
t1 = 273; #inital temperature in K
dv = 2; #ratio of inital volume to final volume
y = 1.4; #coefficient of expansion
# Calculations
r = p/(d*t1); #universal gas constant in dynes.cc/gm.K
t2 = round(t1*(dv)**(y-1)); #final temperature in K
w = r*(t2-t1)/(y-1); #work done in adiabatic compression in ergs
# Result
print 'work done in adiabatic compression is %.3e ergs'%(w)
print "Note : answer is different because of rounding error"
```

In [32]:

```
# Variables
m = 5; #mass of air in gm
cv = 0.172; #specific heat at consmath.tant volume cal/gm
dt = 10; #changi in temperature in K
# Calculations
ie = m*cv*dt; #change in internal energy in cal
# Result
print 'change in internal energy is %3.2f cal'%(ie)
```

In [36]:

```
# Variables
v1 = 10**3; #inital volume in cc
v2 = 2*v1; #final volume in cc
p1 = 76*13.6*981; #pressure in dyne/sq.cm
t1 = 273; #intial temperature in K
d = 1.29; #density of the gas gm/lit
cv = 0.168; #specific heat at constant volume in cal/gm
# Calculations
t2 = (v2/v1)*t1; #final temperature in K
r = 0.068; #universal gas consmath.tant in cal
cp = cv+r; #specific heat at constant pressure in cal
q = d*cp*(t2-t1); #heat supplied in cal
# Result
print 'the heat supplied to the gas is %3.2f cal'%(q)
print "Note: answer is slightly different because of rounding error."
```

In [38]:

```
# Variables
t = 303; #temperature of the one mole of the argon in K
v1 = 1; #intial volume in litres
v2 = 10; #final volume in litres
r = 8.31*10**7; #universal gas constant in ergs/K.mol
# Calculations
w = r*t*math.log(v2/v1); #work done in isothermal expansion in ergs
# Result
print 'the work done in isothermal expansion is %.1e ergs'%(w)
```

In [40]:

```
# Variables
dv = 4; #final volume of neon in lit
t = 273; #temperature of the gas in K
n = 2.6/22.4; #the no.of moles of neon
r = 1.98; #universal gas constant in cal/K.mol
# Calculations
w = n*t*r*math.log(dv); #work done by gas in ergs
# Result
print 'the work done by 2.6lit of neon is %3.2f ergs'%(w)
print "Note: answer is slightly different because of rounding error."
```

In [46]:

```
import math
# Variables
dv = 10**(-3); #ratio of initial and final volume
t1 = 10**5; #initial temperature in K
y = 1.66; #coefficient of expansion
# Calculations
t2 = t1*((4./3*math.pi*10**12)/(4./3*math.pi*10**15))**(y-1); #final temperature in K
# Result
print 'final temperature of the gas is %3.2f K'%(t2)
print "Note : Answer in book is wrong. Please calculate manually."
```

In [42]:

```
# Variables
p1 = 8.; #intial pressure in cm of hg
p2 = 6.; #final pressure in cm of hg
v1 = 1000.; #intial volume in cc
v2 = 1190.; #final volume in cc
# Calculations
y = math.log(p1/p2)/math.log(v2/v1); #coefficient of expansion
# Result
print 'the coefficent of expansion is %3.2f'%(y)
```