Important First Law of Thermodynamics Formulas for JEE and NEET

 Page 1


Important Formulae 
1.  Thermodynamics 
(i) Molar heat capacity C = heat required to raise the temperature of 1 mole of any substance by 
1 °C or 1 K. 
   
Q
nT
?
?
 
   Q = nC ?T 
Molar heat capacity of solids and liquids is almost constant. In case of gases C is process 
dependent. It varies from 0 to ?. In isothermal process, 
   C = ? as ?T = 0 
  In adiabatic process, 
   C = 0 as Q = 0 
C
p
 (molar heat capacity of isobaric process) and C
v
 (molar heat capacity of isochoric process) are 
commonly used. In a general process pV
x
 = constant, molar heat capacity is given by, 
   
RR
C
1 1 x
??
? ? ?
 
(ii) First law of thermodynamics It is a law of conservation of energy given by, 
   Q = ?U + W 
(iii) Detailed discussion of three terms of first law of thermodynamics 
(a) Work done Following methods are generally used to find the work done, 
Method 1 
i
i
V
V
W pdV ?
?
 (because dW = pdV) 
Here, p should be either constant or function of V. If p is constant. It means process is isobaric, 
   W = p(V
f
 – V
i
) = p ?V 
Method 2 Work done can also be obtained by area under p-V diagram with projection on V-axis. 
 
 
W = + ve as volume is increasing 
 
W =
 
- ve as volume is decreasing 
Page 2


Important Formulae 
1.  Thermodynamics 
(i) Molar heat capacity C = heat required to raise the temperature of 1 mole of any substance by 
1 °C or 1 K. 
   
Q
nT
?
?
 
   Q = nC ?T 
Molar heat capacity of solids and liquids is almost constant. In case of gases C is process 
dependent. It varies from 0 to ?. In isothermal process, 
   C = ? as ?T = 0 
  In adiabatic process, 
   C = 0 as Q = 0 
C
p
 (molar heat capacity of isobaric process) and C
v
 (molar heat capacity of isochoric process) are 
commonly used. In a general process pV
x
 = constant, molar heat capacity is given by, 
   
RR
C
1 1 x
??
? ? ?
 
(ii) First law of thermodynamics It is a law of conservation of energy given by, 
   Q = ?U + W 
(iii) Detailed discussion of three terms of first law of thermodynamics 
(a) Work done Following methods are generally used to find the work done, 
Method 1 
i
i
V
V
W pdV ?
?
 (because dW = pdV) 
Here, p should be either constant or function of V. If p is constant. It means process is isobaric, 
   W = p(V
f
 – V
i
) = p ?V 
Method 2 Work done can also be obtained by area under p-V diagram with projection on V-axis. 
 
 
W = + ve as volume is increasing 
 
W =
 
- ve as volume is decreasing 
 
W = 0 as volume is constant 
 
W = +ve as cyclic process is clockwise with p on y-axis.  
(b) Change in internal energy ?U 
  ?U = nC
V
?T
 
for all processes.  
For this C
V
 (or nature of gas), n and ?T should be known.  
  If either of the three terms is not known, we can calculate ?U by, 
   ?U = Q - W 
  (c) Heat exchange Q 
    Q = nC ?T.  
  For this n, ?T and molar heat capacity C should be known. C is a process dependent. So, 
  if either of the three terms (n, ?T or C) is not known, we can calculate Q by, 
    0 = ?U + W  
  (iv) 
dp p
x
dV V
?? in process pV
x
 = constant 
or  slope of p-V graph = 
p
x
V
? 
  In isobaric process x = 0, therefore slope = 0 In isothermal process x = 1, therefore slope = 
p
V
? 
  In adiabatic process x = ?, therefore slope = 
p
V
??