Question

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 1

Let `n_(1)` and `n_(2)` donote the number of moles of gas in the large and small bulbs, in the final configuration, respectively. Denoting the final temperature by `T_(1)` and `T_(2)` and the final pressure by `P_(f)` the ideal gas law implies that
`p_(f) V_(1) = n_(1) RT_(1)`
and `p_(1) V_(2) = n_(2) RT_(2)`
Where `p_(0), V_(0) = V_(1) + V_(2)` and `T_(0)` are the initial pressure, volume, and temperature, respectively.
Using Eqs. (i) and (ii) in the equation `n_(1) + n_(2) = n`, we get
`(p_(f) V_(1))/(RT_(1)) + (p_(f) V_(2))/(RT_(2)) = (p_(0) V_(0))/(RT_(0))`
Solving for `P_(f)`, we obtain `p_(f) = (p_(0) V_(0))/(T_(0) ((V_(1))/(T_(1)) + (V_(2))/(T_(2))))`
Inserting the numerical value, `p_(0) = 1.00 atm, T_(0) = 293.15`
`K (t_(0) = 20^(@) C)`,
`T_(1) = 373.15 K (t_(1) = 100^(@) C)`, and `T_(2) = 237.15 K`
`(t_(2) = 0^(@) C)`,
we find `p_(f) = ((1.00)(600))/((293.15) [((400)/(373.15)) + ((200)/(273.15))]) = 1.13` atm