Question 243: Thermodynamics & KTG

Question

The pressure and volume of an ideal gas are related as $\mathrm{PV}^{\frac{3}{2}}=\mathrm{K}$ (Constant). The work done when the gas is taken from state $A\left(P_1, V_1, T_1\right)$ to state $B\left(P_2, V_2, T_2\right)$ is :

Accepted Answer

To find the work done by the gas when it goes from state A to state B, we can look at the definition of work done on or by a gas in a thermodynamic process. For a quasi-static process, the work done $$ W $$ is given by the integral of the pressure $$ P $$ with respect to the volume $$ V $$:
$$ W = \int_{V_1}^{V_2} P dV $$
Given the relationship $$ PV^{\frac{3}{2}} = K $$, we can solve for $$ P $$:
$$ P = \frac{K}{V^{\frac{3}{2}}} $$
Now, substitute $$ P $$ into the work integral and evaluate it:
$$ W = \int_{V_1}^{V_2} \frac{K}{V^{\frac{3}{2}}} dV = K \int_{V_1}^{V_2} V^{-\frac{3}{2}} dV $$
To integrate this, we'll use the power rule for integration. The integral of $$ V^{-\frac{3}{2}} $$ is $$ -2V^{-\frac{1}{2}} $$, so the work done is:
$$ W = K \left[-2V^{-\frac{1}{2}}ight]_{V_1}^{V_2} = K \left(-2V_2^{-\frac{1}{2}} + 2V_1^{-\frac{1}{2}} ight) $$
We can rewrite $$ V^{-\frac{1}{2}} $$ as $$ \frac{1}{\sqrt{V}} $$:
$$ W = K \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} ight) $$
Since $$ P_2V_2^{\frac{3}{2}} = K $$ and $$ P_1V_1^{\frac{3}{2}} = K $$, we can express $$ K $$ in terms of $$ P_1 $$ and $$ V_1 $$ (or $$ P_2 $$ and $$ V_2 $$, but weÃ¢ÂÂll use $$ P_1 $$ and $$ V_1 $$ for now):
$$ W = P_1V_1^{\frac{3}{2}} \left(-2\frac{1}{\sqrt{V_2}} + 2\frac{1}{\sqrt{V_1}} ight) $$
$$ W = -2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_2}} + 2P_1V_1\sqrt{V_1}\frac{1}{\sqrt{V_1}} $$
$$ W = -2P_1V_1\frac{\sqrt{V_1}}{\sqrt{V_2}} + 2P_1V_1 $$
Since $$ \frac{\sqrt{V_1}}{\sqrt{V_2}} = \sqrt{\frac{V_1}{V_2}} $$, and recalling $$ P_2V_2^{\frac{3}{2}} = K $$ once more:
$$ W = -2P_1V_1\sqrt{\frac{V_1}{V_2}} + 2P_1V_1 $$
$$ W = -2 \frac{P_1V_1^{\frac{3}{2}}}{\sqrt{V_2}} + 2P_1V_1 $$
Putting $$ P_1V_1^{\frac{3}{2}} $$ back as $$ K $$:
$$ W = -2\frac{K}{\sqrt{V_2}} + 2P_1V_1 $$
$$ W = -2P_2V_2 \sqrt{V_2}\frac{1}{\sqrt{V_2}} + 2P_1V_1 $$
$$ W = -2P_2V_2 + 2P_1V_1 $$
So the work done is:
$$ W = 2P_1V_1 - 2P_2V_2 $$
Therefore, the correct answer matching the given options is:
Option D: $$ 2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2ight) $$
Shortcut :
For $\mathrm{PV}^{\mathrm{x}}=$ constant
If work done by gas is asked then
$$
\begin{aligned}
\mathrm{W} & =\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\mathrm{x}} \\
& 	ext { Here } \mathrm{x}=\frac{3}{2} \\
	herefore \mathrm{W} & =\frac{\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1}{-\frac{1}{2}} \\
= & 2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2ight)
\end{aligned}
$$

Answer

$2\left(\mathrm{P}_2 \sqrt{\mathrm{V}_2}-\mathrm{P}_1 \sqrt{\mathrm{V}_1}\right)$

Answer

$2\left(\sqrt{\mathrm{P}_1} \mathrm{~V}_1-\sqrt{\mathrm{P}_2} \mathrm{~V}_2\right)$

Answer

$2\left(\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1\right)$

Answer

$2\left(\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2\right)$

Thermodynamics & KTG

The pressure and volume of an ideal gas are related as \mathrm{PV}^{\frac{3}{2}}=\mathrm{K} (Constant). The work done when the gas is taken from state A\left(P_1, V_1, T_1\right) to state B\left(P_2, V_2, T_2\right) is :

Choose the correct answer: