Obtain expressions for the work done, change in internal energy and he

1.	Obtain expressions for the work done, change in internal energy and heat supplied in an isobaric process on a gas
Answer» ${\boxed{\underline{\tt{\orange{Required \: \: answer:-}}}}}$ ★TO FIND:- An expression for work done, CHANGE in internal energy and heat supplied in an isobaric process on a gas ★SOLUTION:- Isobaric process MEANS a thermidynamic process ASSOCIATED with CONSTANT external pressure Let, Internal energy change = ∆U Number of moles = μ Work = W Heat change = ∆Q In Constant Pressure, the heat change is :- $: \longrightarrow \displaystyle \sf{∆Q = \mu C_p\triangle\theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)}$ In all thermodynamics process, ∆U is given as :- $: \longrightarrow \displaystyle \sf{∆U = \mu C_v \triangle\theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2)}$ As according to first law of thermodynamics:- $\displaystyle \sf{ \therefore∆Q = ∆U + W} \\ \\ \rightarrow \displaystyle \sf{ \mu C_p \triangle\theta = \mu \: C_v \triangle \theta + W} \\ \\ { \boxed{ \sf{W = \mu \: (C_v - C_p) \triangle \: \theta}}}$ It can also be represented by:- $\rightarrow \displaystyle{ \boxed{ \sf{ \pink{W = P∆V}}}}$

Obtain expressions for the work done, change in internal energy and heat supplied in an isobaric process on a gas

Answer»

${\boxed{\underline{\tt{\orange{Required \: \: answer:-}}}}}$

★TO FIND:-

An expression for work done, CHANGE in internal energy and heat supplied in an isobaric process on a gas

★SOLUTION:-

Isobaric process MEANS a thermidynamic process ASSOCIATED with CONSTANT external pressure

Let,

Internal energy change = ∆U

Number of moles = μ

Work = W

Heat change = ∆Q

In Constant Pressure, the heat change is :-

$: \longrightarrow \displaystyle \sf{∆Q = \mu C_p\triangle\theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)}$

In all thermodynamics process, ∆U is given as :-

$: \longrightarrow \displaystyle \sf{∆U = \mu C_v \triangle\theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2)}$

As according to first law of thermodynamics:-

$\displaystyle \sf{ \therefore∆Q = ∆U + W} \\ \\ \rightarrow \displaystyle \sf{ \mu C_p \triangle\theta = \mu \: C_v \triangle \theta + W} \\ \\ { \boxed{ \sf{W = \mu \: (C_v - C_p) \triangle \: \theta}}}$

It can also be represented by:-

$\rightarrow \displaystyle{ \boxed{ \sf{ \pink{W = P∆V}}}}$