Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0oC and a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in water bath maintained at 62oC. What is the new value of the pressure (in cm of Hg) inside the bulbs? The volume of the connecting tube is negligible.

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answer is 83.75.

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Detailed Solution

If we assume the volume of each bulbs is V, the number of moles present in each bulb initially is n=PVRT=(76cm)VR(273)Thus total number of moles in the two connected bulbs is 2n=2×(76cm)VR(273)when one bulbs is placed in melting ice (273 K) and other is placed in water at 62oC (62 + 273 = 335 K) still total number of moles in the two bulbs will be equal to the initial moles i.e. 2n. If Pf be the final pressure in the two bulbs we havePi(2V)Ti=PfVTf1+PfVTf2(76cm)(2V)273=PfV273+PfV335 or Pf=2×76273×273×335608=83.75cm of Hg

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