91Ó°ÊÓ

If during a process the temperature remains conserved, then the system is known as an isothermal process. It is generally conducted very slowly, so that heat gets ample time to transfer between the system and surroundings. The system undergoing an isothermal process must satisfy the relation-

.$\mathbf{PV}\mathbf{=}\mathbf{constant}$

The work done by the system during an isothermal process is given as-

$\mathbf{W}\mathbf{=}\mathbf{nRT}\text{ }\mathbf{ln}\left(\frac{{\mathbf{V}}_f}{{\mathbf{V}}_i}\right)$

Here, P,V and T are pressure, volume and temperature respectively and R is the gas constant.

1. The initial volume of an ideal gas is,${\text{V}}_i=4.00{\text{″¾}}^3$
2. The final volume of an ideal gas is,${\text{V}}_f=3.00{\text{m}}^3$
3. The number of moles of an ideal gas,$\text{n}=3.50$
4. Temperature of gas is,$\text{T}=10°\text{C}=283\text{K}$

The equation for work done by an ideal gas is

$\text{W}=\text{²Ô¸é°Õ l²Ô}\left(\frac{{\text{V}}_f}{{\text{V}}_i}\right)$

$\text{W}=\left(3.50\text{″¾ol}\right)×(8.31\text{ J/³¾´Ç±ô.°­})×\left(283\text{‿é}\right)×\ln \left(\frac{{\displaystyle 3.00{\text{″¾}}^3}}{{\displaystyle 4.00{\text{″¾}}^3}}\right)$

$\text{W}=−2.37×{10}^3\text{J}$

The equation of the first law of thermodynamics is

$\mathrm{Δ}{\text{E}}_{int}=\text{Q}−\text{W}$

When the temperature of the gas remains constant, the internal energy also remains constant.

So,$\mathrm{Δ}{\text{E}}_{int}=0$. Using this, in the above equation, we get

$\begin{array}{c}\text{Q}=\text{W}\\ =−2.37×{10}^3\text{J.}\end{array}$