Physics, asked by vikas33236, 2 months ago

What is an isothermal process
Obtain an expression for work done by the gas during this process. ​

Answers

Answered by BrainlyTwinklingstar
9

Isothermal process is a system which undergoes physical changes where the temperature remains constant is known as isothermal process.

As the temperature is constant, the relationship between pressure and volume of the gas is given by,

PV = K (constant)

If the gas expands through small volume dV , Workdone by the gas,

\dashrightarrow \sf dW = PdV

Now, integrate the equation between V₁ and V₂

\sf\dashrightarrow W = \int \limits_{v_1} ^{v_2} PdV

From ideal gas equation,

PV = RT

P = RT/V

\sf\dashrightarrow W = \int \limits_{v_1} ^{v_2}  \dfrac{RT}{V}dV

\sf\dashrightarrow W =RT \int \limits_{v_1} ^{v_2}  \dfrac{dV}{V}

\sf\dashrightarrow W =RT \int \limits_{v_1} ^{v_2}  \dfrac{1}{V}dV

\sf\dashrightarrow W =RT \bigg[LogV\bigg]_{v_1} ^{v_2}

\sf\dashrightarrow W =RT  \big(log(V_2) - log(V_1) \big)

\sf\dashrightarrow W =RT log_e \bigg( \dfrac{V_2}{V_1}  \bigg)

\sf\dashrightarrow W =2.303RT \:  log_{10}\bigg( \dfrac{V_2}{V_1}  \bigg)

we know that,

 \sf P_1V_1 = P_2V_2

 \sf  \dfrac{V_2}{V_1} =  \dfrac{P_1}{P_2}

\dashrightarrow  \underline{\boxed{ \sf W = 2.303RT \:  log_{10}\bigg( \dfrac{P_1}{P_2}  \bigg)}}

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