A glass tube of uniform area of cross section and open at one end encloses some air at 27^@C by a 4 cm long mercury thread that acts like a piston. When the tube is held vertical with its open end up length of the air column in the tube is 9 cm.When the open end is held downwards by turning the tube,the length of the enclosed air column becomes 10 cm. Find () the value of the atmospheric pressure (ii) the temperature at which the length of the air column becomes 9cm again,while the tube is still held inverted.

A glass tube of uniform area of cross section and open at one end encl

1.	A glass tube of uniform area of cross section and open at one end encloses some air at 27^@C by a 4 cm long mercury thread that acts like a piston. When the tube is held vertical with its open end up length of the air column in the tube is 9 cm.When the open end is held downwards by turning the tube,the length of the enclosed air column becomes 10 cm. Find () the value of the atmospheric pressure (ii) the temperature at which the length of the air column becomes 9cm again,while the tube is still held inverted.
Answer» SOLUTION :Let the ATMOSPHERIC pressure =pcmHg and the area of cross section of the tube =`a cm^2`[Fig.6.10] when the open end is kept upwards the VOLUME of the confined air `V_1=9 a cm^3` and pressure `p_1=(p+4) cmHg`. On INVERTING the tube volume of the confined air `V_2=10 a cm^3` and pressure `p_2(p-4)`cmHg. Assuming temperature to be constant at `27^@C` from Boyle.s law `p_1V_1=p_2V_2` or,`9a(p+4)=10a(p-4)` or,`9p+36=10p-40 , or p=76 cmHg` Suppose at `T_2K` the length of the air column becomes 9 cm again when the tube is still held inverted.Then it pressure `p_2=76-4=72 cmHg` and volume `V_2=9a cm^3`. `therefore(p_1V_1)/T_1=(p_2V_2)/T_2` or,`((76+4)times9a)/(273+27)=(72times9a)/T_2or,80/300=72/T_2` `thereforeT_2=270K=(270-273)^@C=-3^@C`