Question

define the molar heat capacities Cp and Cv for a gas.Show that, for a mole of an ideal gas, Cp-Cv = R
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Answer 1

$\\ \: \\ \large \sf \underbrace{ \underline{Understanding \: concept :- }}$

In this , Where Cp represent the specific heat at pressure ; dH is the change in enthalpy; dT is the change in the temperature. Cv: During a small change in the temperature of a substance, Cv is the amount absorbed/released per unit mass of a substance where volume does not change.

Now,

$\\ \large \underline { \underline{ \sf \: Proof :- }}$

We have to show that Cp - Cv = R

So , let's start

According to first law of thermodynamics :

∆Q = ∆U + ∆W where, ∆Q is the amount of heat that is given to the system, ∆U is the change in internal energy and ∆W is the work done.

We can write :

∆Q = ∆U + P∆U , as ∆W = P∆V

Since, ∆Q =$\sf nC_p∆T$ and ∆U = $\sf nC_vC∆T$

Therefore, $\sf nC_p∆T$= $\sf nC_v∆T$ + P∆ ------------------- (1)

We know that, PV = nRT

At $\sf{T_1}$ Kelvin = $\sf PV_1$ = $\sf nRT_1$ ------------------ (a)

At $\sf{T_2}$ Kelvin = $\sf PV_2$ = $\sf nRT_2$ ------------------ (b)

Subtracting a from b : so we get ,

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: PV_2 - PV_1 = nRT_2 - nRT_1$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: P(V_2 - V_1 )= nR(T_2 - T_1)$

Where, $\sf V_2 - V_1$= ∆V and $\sf T_2 - T_1$∆T

Therefore, P∆V = nR∆T

Putting the values of P∆V in equation (1):

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: nC_p∆T \: = \: nC_v∆T + nR∆T$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \sf \: nC_p∆T \: = \: n∆T (C_v+ R)$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \sf \: C_p \: = C_v+ R$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: \sf \: C_p \: - C_v = R$

Hence proved✓