Question

An ideal gas is taken from an initial state i to a final state f in such a way that the ratio of the pressure to the absolute temperature remains constant. What will be the work done by the gas?

Answer 1

The work done by the gas is $p \Delta V=0$

Explanation:

$P_1$ is Initial pressure

$P_2$ is  Final pressure

$T_1$ is  Absolute initial temperature

$T_2$ is  Absolute final temperature

$\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}$

Step 1:

Using the ideal equation for gas we get

Where V is the volume  

If n is considered as the number of gas moles and R is the constant of universal gas, then

Step 2:

$\frac{P_{1}}{T_{2}}=\frac{n R}{V_{1}}$      

and $\frac{P_{2}}{T_{2}}=\frac{n R}{V_{2}}$

$\begin{array}{l}{V_{1}=V_{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots\left[\frac{P_{1}}{T_{2}}=\frac{P_{2}}{T_{2}}\right]} \\{\Delta V=V_{2}-V_{1}=0}\end{array}$

$Thus, Gas Work done =p \Delta V=0$

Answer 2

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Delta P =0