Question

The specific heat capacity of a monatomic gas for the process tv2 = constant is

Answer 1

Given:

Monoatomic gas follows the process TV² = constant.

To find:

Specific heat capacity for this process.

Calculation:

$\therefore \: T {V}^{2} = constant = k$

$= > \: \dfrac{PV}{nR} \times {V}^{2} = k$

$= > \: \dfrac{P{V}^{3} }{nR} = k$

$= > \: P{V}^{3}= c$

Let c be another constant.

For these Thermodynamic Process, where $\boxed{\sf{P{V}^{N} = constant }}$;

Let C be specific heat capacity:

$\therefore \: C = C_{V} + \dfrac{R}{1 - N}$

$= > \: C = \dfrac{3R}{2} + \dfrac{R}{1 - 3}$

$= > \: C = \dfrac{3R}{2} - \dfrac{R}{2}$

$= > \: C = R$

So, final answer is:

$\boxed{ \red{ \bold{ \large{\: C = R}}}}$

Answer 2

Given:

A monatomic gas for the process tv2 = constant

To find:

The specific heat capacity of a monatomic gas for the process tv2 = constant is

Solution:

From given, we have,

A monatomic gas for the process tv2 = constant

Let us consider,

⇒ TV² = C

differentiating the above equation, we get,

dV/dT = - V/2T ........(1)

Using the law of thermodynamics, we have,

dQ = dU + dW

⇒ C = Cv + PdV/ndT .......(2)

Using the ideal gas law, we have,

PV = nRT ........(3)

⇒ P/n = RT/V .....(4)

consider the equation (3), we have,

C = Cv + PdV/ndT

⇒ C = Cv + (P/n) × (dV/dT) .....(5)

substitute the equations (1) and (4) in (5), we get,

C = R/2

Therefore, the specific heat capacity of a monatomic gas for the process tv2 = constant is R/2