Question

obtain an expression for work done by the gas during an adiabatic process


Please answer as soon as possible​

Answer 1

In adiabatic process PV$\sf ^{\gamma}$ is constant.

If 'w' is the Workdone during adiabatic process and if the gas expands through small volume dv, work done by the gas is

$\sf\dashrightarrow dW = Pdv$

Now, integrate the equation between V₁ and V₂

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

we know that,

PV$\sf ^{\gamma}$ = K

P = KV$\sf ^{-\gamma}$

$\sf\dashrightarrow W =K \int \limits_{v_1} ^{v_2} V ^{-\gamma} dv$

$\sf\dashrightarrow W =K \bigg( \dfrac{ {V}^{ - \gamma + 1} }{1 - \gamma} \bigg) _{v_1} ^{v_2}$

$\sf\dashrightarrow W =K \bigg( \dfrac{ {V_{2}}^{1 - \gamma } }{1 - \gamma} - \dfrac{ {V_{1}}^{ 1- \gamma} }{1 - \gamma} \bigg)$

$\sf\dashrightarrow W = \dfrac{1 }{1 - \gamma} [K{V_{1}}^{ - \gamma + 1} - K{V_{1}}^{ - \gamma + 1}]$

$\sf \: P_1V_1^\gamma = P_2V_2^\gamma = K$

$\sf\dashrightarrow W = \dfrac{1 }{1 - \gamma} [P_1V_1^\gamma {V_{1}}^{ - \gamma + 1} - P_2V_2^\gamma {V_{1}}^{ - \gamma + 1}]$

$\sf\dashrightarrow W = \dfrac{1 }{1 - \gamma} [P_1 {V_{1}} - P_2V_2]$

From ideal gas equation,

$\sf PV = RT$

$\sf P_1V_1 = RT_1$

$\sf P_2V_2 = RT_2$

$\sf\dashrightarrow W = \dfrac{1 }{1 - \gamma} [RT_1 - RT_2]$

$\sf\dashrightarrow W = \dfrac{R}{1 - \gamma} [T_1 - T_2]$

For n moles of a gas,

$\dashrightarrow \underline{ \boxed{ \sf W = \dfrac{nR}{1 - \gamma}[T_1 - T_2]}}$