Question

16.
A piston is cleverly designed so that it extracts the maximum amount of work out of a chemical
reaction, by matching Pexternal to the Pinternal at all times. This 8 cm diameter piston initially holds back
1 mol of gas occupying 1 L, and comes to rest after being pushed out a further 2 L at 25°C. After
exactly half of the work has been done, the piston has travelled out a total of
(A) 10.0 cm (B) 11.2 cm (C) 14.5 cm (D) 20.0 cm​

Answer 1

Answer:

option C is correct.

Explanation:

first we have to find the radius by using given diameter value

r=$\frac{d}{2} =\frac{8}{2} =4cm$

n=1 mole

$V_{1} =1Lit$

$V_{2} =2Lit$

T=25C

Here $P_{net} =P_{internal}$

so here the process is isothermal reversible

W=-2.303nRT$log\frac{V_{2} }{V_{1} }$

W=-2.303*1*8.314*298log2

$W_{half} =\frac{W}{2} =-nRTlog(\frac{V_{f} }{V_{1} } )*2.303$

=$\frac{-2.303*1*8.314*298*log2}{2}$

=-2.303*1*8.314*298*log$\frac{V^{'} }{2}$

so we get the value of $V^{'}$ is $2\sqrt{2} Lit$

Volume=area*length

$V_{2} =A*l_{2}$

⇒$2*10^{3} =\frac{22}{7} *(4)^{2} *l_{2}$

$l_{2} =\frac{14000}{352}$$V^{'} =A*l^{'}$

⇒$2*1.4*10^{3} =\frac{22}{7} *(4)^{2} l^{'}$

$l^{'} =\frac{19000}{352}$

Distance travelled by the piston is Δl=$\frac{19000-14000}{352}$

Δl=$\frac{500}{352}$

=14.5cm

option C is correct