Question

Question: In the temperature range 310 K to 330 K, The pressure P of a certain non ideal gas is related to volume V and temperature T by$\mathit{p}\mathbf{=}\left(24.9\frac{J}{K}\right)\frac{\mathbf{T}}{\mathbf{V}}\mathbf{-}\left(0.00662\frac{J}{K^2}\right)\frac{{\mathbf{T}}^{\mathbf{2}}}{\mathbf{V}}$

How much work is done by the gas if its temperature is raised from 315 Kto 325 Kwhile the pressure is held constant?

Short answer

Answer

Work done by the gas is 207 J.

Step-by-step solution

Step 1: Given

$$\begin{gathered} T_1=315\text{K} \\ {T}_2=325\text{K} \\ P=\left(24.9\right)\frac{{\displaystyle T}}{{\displaystyle V}}-\left(0.00662\right)\frac{{\displaystyle T^2}}{{\displaystyle V}} \end{gathered}$$

Determine the concept

Formula is as follow:

$\mathbf{W}\mathbf{=}{\mathbf{P}}_{\Delta }\mathbf{V}$

Here, Pis pressure, V is volume and W is work done.

Determine the work done by the gas

Write the expression for pressure and volume as:

$$\begin{gathered} P=\left(24.9\right)\frac{T}{V}-\left(0.00662\right)\frac{T^2}{V} \\ \therefore V=\left(24.9\right)\frac{T}{P}-\left(0.00662\right)\frac{T^2}{P} \end{gathered}$$

Solve for the work done as:

$$\begin{gathered} W=P▵V \\ W=P\left(\left(24.9\right)\frac{T}{P}-\left(0.00662\right)\frac{T^2}{P}\right) \\ W=P\left(\left(24.9\right)\frac{▵T}{P}-\left(0.00662\right)\frac{▵\left(T^2\right)}{P}\right) \\ W=P\left(\left(24.9\right)\frac{\left(T_2-T_1\right)}{P}-\left(0.00662\right)\frac{T_2^2-T_1^2}{P}\right) \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} W=P\left(\left(24.9\right)\frac{\left(325-315\right)}{P}-\left(0.00662\right)\frac{({325}^2-{315}^2)}{P}\right) \\ W=\left(\left(24.9\right)\left(325-315\right)-\left(0.00662\right)({325}^2-{315}^2)\right) \\ W=206.632\approx 207\text{J} \end{gathered}$$

Hence, thework done by the gas is 207 J.

Therefore, the work done by the gas can be found by using the formula for work done in isobaric process.