please explain__ cp-cv=R __ briefly

Relationship between C_{v} and C_{p} for an ideal gas is as follows:

Heat capacity, C=q/T [where, q: Heat; T: Temperature]

**At constant volume**, Heat capacity is denoted by C_{v}.

So, C_{v}= q_{v}/∆T

Or, q_{v}=C_{v}∆T -----------------(1)

Since, ∆U=q - P∆V [Where, ∆U: Internal energy change]

At constant volume , P∆V=0

Therefore, at constant volume, ∆U=q_{v} -------(2)

From (1) and (2),

**∆U =C _{v}∆T**

**At constant pressure**, Heat capacity is denoted by C_{p}.

So, C_{p}= q_{p}/∆T

q_{p}= C_{p}∆T ---------------------(3)

we know, ∆H=∆U+∆(PV) [where, ∆H:enthalpy change]

At constant pressure, ∆H=∆U+ P∆V

Since at constant pressure, ∆U=q_{p}- P∆V

q_{p}=∆U+P∆V

q_{p}=∆H -------------(4)

From (3) and (4),

**∆H= C _{p}∆T**

Since, ∆H=∆U+∆(PV)

∆H=∆U+∆(nRT) [we are deriving Relationship between C_{v} and C_{p} for an ideal gas]

∆H=∆U+∆(RT) [For 1 mole of an ideal gas]

∆H=∆U+R∆T

After putting the values of ∆H and ∆U in above equation ,we get,

C_{p}∆T = C_{v}∆T + R∆T

**C _{p} = C_{v} + R or C_{p}- C_{v} = R**

**
**