A glass bulb of volume 400cm cubed is connected to another of volume 200cm cubed by a tube of negligible volume. Both bulbs initially contain dry air at 20 degrees celsius and 1 atm. The larger bulb is then immersed in steam at 100 degrees celsius and the smaller in melting ice at 0 degrees celsius. Find the common pressure in the system.

Hm.. I suggest using the ideal gas equation, like this:

P_h = \frac{n_hRT_h}{V_h}

P_c = \frac{n_cRT_c}{V_c}

h means hot, and c means cold.

You can cross out R and equate both, since the pressure is the same.

\frac{n_hT_h}{V_h} = \frac{n_cT_c}{V_c}

And from the first equation, you can work out the total number of moles:

n = \frac{PV}{RT}

and n_h + n_c = n

So, n_h + n_c = \frac{PV}{RT}

Thank you for solving this problem.
So firstly I worked out the total number of moles (n) before the bulbs were immersed, which gave me 0.025 moles.
Then, I substituted n - n(c) for n(h) in the first equation (after the immersion) and then was able to work out n(c), since I then only had one unknown (after crossing out R and equating both). This gave me 0.01 moles for n(c). Then I worked out the pressure for n(h) x T(h) / V(h), which gave me a pressure after the immersion of 13,850 Pascals. Hopefully, this gives me the right answer. I presume that the temperature of the 2 bulbs will equilibrate a short time after and T(h) and T(c) will be the same and the answer of 13,850 Pascals will still hold. Also, since the pressure before the immersion was 1 atmosphere (101,325 Pascals) and is now only 13,850 Pascals after the immersion, then the overall temperature has decreased from 20 degrees celsius (before the immersion), since temperature decreases with decreasing pressure.

Wait... how did you get 13850 Pa? I'm not getting that whatever I do.

Though I agree with the other values you got :)

And thanks for the thumb up :)

Thanks for getting back again.
So I got approximately 0.01 moles for n(c). Also since n = n(h) + n(c), then n(h) = n - n(c), which gives approximately 0.015 moles (actually, it works out at 0.0149 moles, I think). So, using the left side of the equation for the hot part, P(after) = n(h) x T(h) / V (h) = (0.0149 moles x 373 Kelvin) / 400 x 10 -6 (to the minus 6) metres cubed = 13,800 Pascals (approximately). Similarly, using using the right side of the equation for the cold part, P(after) = n(c) x T(c) / V (c) = (0.0101 moles x 273 Kelvin) / 200 x 10 -6 (to the minus 6) metres cubed = 13,800 Pascals (approximately). I hope my units are correct.

Oh, you forgot the ideal gas constant which I crossed out to get my formula, look back at the very first formulae :)

Otherwise, to have currently the value of P/R

Sorry, I wasn't concentrating fully when working out the answer.
So putting in the gas constant (R = 8.314) gives me the pressure for the hot side (after immersion in steam at 100 degrees celsius), which is P(hot) = ( n(h) x R x T(h) ) / V(h) = (0.0149 moles x 8.314 x 373 Kelvin) / 400 x 10 -6 (to the minus 6) = 115,500 Pascals approximately. So the pressure before the immersion was 101, 325 Pascals and the Pressure after the immersion was 115,500 Pascals. This probably makes more sense, since pressure increases with increasing temperature and two thirds of the system (400 x 10 -6 metres cubed) had been immersed in steam at 100 degrees celsius.
Thanks again. This is a helpful site. I will recommend it.

Yes, that's it well done Sean :)