(i) A conducting cylinder whose inside diameter is 4.00cm contains air compressed by a piston of mass m = 13.0 kg, which can slides freely in the cylinder shown in the figure. The entire arrangement is immersed in a water bath whose temperature can be controlled. The syetem is initially in equilibrium at temperature t_(1) = 20^(@)C. The initial height of the piston above the bottom of the cylinder is h_(i) = 4.00 cm. P_(atm) =1 xx 10^(5) N/m^(2) and g = 10m//s^(2). If the temperature of the water bath is gradually increased to a final temperature t_(f) = 100^(@)C. Find teh height h_(f) of the piston (in cm) at that instant? (ii) In the above question, if we again start from the initial conditions and the temperature is again gradually raised, and weights are added to the piston to keep its height fixed at h_(i). Find the value of teh added mass when the final temperature becomes t_(f) = 100^(@)C?

(i) A conducting cylinder whose inside diameter is 4.00cm contains air

1.	(i) A conducting cylinder whose inside diameter is 4.00cm contains air compressed by a piston of mass m = 13.0 kg, which can slides freely in the cylinder shown in the figure. The entire arrangement is immersed in a water bath whose temperature can be controlled. The syetem is initially in equilibrium at temperature t_(1) = 20^(@)C. The initial height of the piston above the bottom of the cylinder is h_(i) = 4.00 cm. P_(atm) =1 xx 10^(5) N/m^(2) and g = 10m//s^(2). If the temperature of the water bath is gradually increased to a final temperature t_(f) = 100^(@)C. Find teh height h_(f) of the piston (in cm) at that instant? (ii) In the above question, if we again start from the initial conditions and the temperature is again gradually raised, and weights are added to the piston to keep its height fixed at h_(i). Find the value of teh added mass when the final temperature becomes t_(f) = 100^(@)C?
Answer» Solution :(i) Initial pressure of gas `P_(i) = P_(atm) +(mg)/(A) =` final pressure `V_(i) = Ah_(i)` and `V_(f) = Ah_(f)` For air inside cylinder `n_(1) = n_(2) rArr (P_(i)V_(i))/(T_(i)) = (P_(f)V_(f))/(T_(f))` `rArr h_(f) = (h_(i)T_(f))/(T_(i)) = (4 XX 373)/(293) = (1492)/(293) cm` (II) Here volume is constant `(P_(i))/(T_(i)) = (P_(f))/(T_(f))` `m = (P_(atm)(T_(f)-T_(i))A+Mg(T_(f)-T_(i)))/(gT_(i))` `= (1 xx 10^(5) (80)pi(2xx10^(-2))^(2)+13(80)xx10)/(10xx293)` `= (320pi)/(293) (1+(13)/(4pi)) KG`