A glass bulb of the 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 . 0 atm , respectively. The larger bulb is immersed in steam at 100 ° C and the smaller in melting ice at 0 ° C . Find the final common pressure.

Step-by-step solution

Let n 1 and n 2 denote the number of moles of gas in the large and small bulbs, in the final configuration, respectively. Denoting the final temperatures 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 ...(i) and p f ⁡ V 2 = n 2 RT 2 ...(ii) where p 0 , V 0 = V 1 + V 2 and T 0 are the initial pressure, volume, and temperature, respectively. p f ⁡ V 1 RT 1 + p f ⁡ V 2 RT 2 = p 0 V 0 RT 0 Solving for Pf, we obtain p f ⁡ = p 0 V 0 T 0 V 1 T 1 + V 2 T 2 Inserting the n