Question

Question: Supposeof an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. If figure shows the final volume V_f versus Q, what is the gas temperature? The scale of the vertical axis is set by ${\mathit{V}}_{fs}\mathbf{=}\mathbf{.}\mathbf{30}{\mathbf{m}}^{\mathbf{3}}$, and the scale of the horizontal axis is set by ${\mathit{Q}}_s\mathbf{=}\mathbf{1200}\mathit{J}$

Short answer

Answer

The gas temperature is 360 K

Step-by-step solution

Step 1: Given

The graph $V_f\text{vs} Q$

$$\begin{gathered} Q_s=1200\text{J} \\ n=0.825\text{mol} \\ {V}_f=0.30{\text{m}}^3 \\ {V}_i=0.20{\text{m}}^3 \end{gathered}$$

Determine the concept

Using that formula for work done by the gas at a constant temperature and substituting the values from the given graph, find the gas temperature.

Formula is as follow:

$W=nRT\ln \left(\frac{V_f}{V_i}\right)$

Here, W is work done, V is volume, T is temperature, R is universal gas constant and n is number of moles.

Determine the gas temperature

For isothermal process,

Q = W

Also,

$W=nRT\ln \left(\frac{V_f}{V_i}\right)$

For isothermal process,

$$\begin{gathered} \therefore Q=nRT\ln \left(\frac{V_f}{V_i}\right) \\ \therefore T=\frac{Q}{nR\ln \left(\frac{V_f}{V_i}\right)} \end{gathered}$$

From the given graph,

Q = 1000 J when$V_{fs}=0.3{\text{m}}^2$ ,

$$\begin{gathered} T=\frac{1000}{0.825\left(8.314\right)\ln \left(\frac{0.30}{0.20}\right)} \\ T=359.56\text{K}\approx 360\text{K} \end{gathered}$$

Hence, the gas temperature is 360 K

Therefore, the temperature of the gas for isothermal process can be found by using the formula for work done by the gas at a constant temperature.