A glass bulb of volume `400 cm^(3)` is connected to another bulb of volume `200 cm^(3)` by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of `20^(@) C` and 1.000 atm, respectively. The larger bulb is immersed in steam at `100^(@) C` and the smallar in melting ice at `0^(@)`. Find the final common pressure.

Answer

A, C

Explanation

`(n_1 +n_2)_1 = (n_1 +n_2)f`
`:. (p_i V_1)/(R T _i) + (p_i V_2)/(RT_i) = (p V_1)/(R T_1)+ (p V_2)/(R T_2)`
`p = (p(V_1 + V_2))/(T_i(V_1//T_1 + V_2//T_2)`
= `((1 a t m)(600))/((293)((400)/(373)+ (200)/(273))`
= `1.13 "a t m"`.