Find the Compression!

Classical Mechanics Level 5

Consider the diagram shown above where the two chambers separated by piston-spring arrangement contain equal amounts of same ideal gas. Initially when the temperatures of the gas in both the chambers are kept at $300 K$ . The compression in the spring is $10 \text{ cm}$ . The temperature of the left and the right chambers are now raised slowly to $400 K$ and $500 K$ respectively. The pistons are free to slide (i.e. no friction between piston and chamber). Final compression in the spring is found to be ( $\frac{\alpha}{60}m$ ). Find $\alpha$ .

Use $\sqrt{481} = 22$ .

The answer is 9.

1 solution

Nishant Rai
May 20, 2015

Let $l_1$ & $l_2$ be the final length of the two parts, then from gas equation,

$\huge \frac{P_0 A l_0}{T_0} = \frac{P_1 A l_1}{T_1} = \frac{P_2 A l_2}{T_2} .......(i)$

Considering the equilibrium of the piston in the initial and the final position,

$\huge P_0 A = kx_0$ & $\huge PA=kx$

$\huge \Rightarrow \frac{P}{P_0} = \frac{x}{x_0}.......(ii)$

Decrease in the length of the spring = Increase in the length of the two chambers

$\huge x - x_0 = l_1 + l_2 - 2l_0 ......(iii)$

from relation (i) $\Rightarrow \huge l_1 = \frac{P_0 T_1 l_0}{PT_0}$ & $\huge l_2 = \frac{P_0 T_2 l_0}{PT_0}$

from relation (ii) $\Rightarrow \huge l_1 = \frac{x_0 T_1 l_0}{xT_0}$ & $\huge l_2 = \frac{x_0 T_2 l_0}{xT_0}$

Putting these in equation (ii) & (iii)

$\huge x-x_0 = \frac{x_0l_0}{xT_0}(T_1 + T_2) - 2l_0$

On solving, we get $\huge x = \frac{3}{20} = \frac{9}{60}$

How did P1 and P2 from equation (i) change into P in equation (ii) and in l1 and l2?

Ankush Patanwal - 2 years, 3 months ago

Since in spring to be in equilibrium, force on both sides must be same, which implies pressure on both sides must be same

Sushank Mishra - 9 months, 1 week ago

Find the Compression!

The answer is 9.

1 solution

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