A horizontal uniform glass tube of 100 cm length, sealed at both ends, contains 10 cm mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively 31 C ∘ and 76 cm of mercury. If the air column at one end is kept at 0 C ∘ and the other end at 273 C ∘ , the pressure of air which is at 0 C ∘ is (in cm of Hg ). Assure no heat flow through mercury

Step-by-step solution

On keeping the temperature of the ends of tube at 0 C ∘ and 273 C ∘ . Applying ideal gas equation p 1 V 1 T 1 = p 2 V 2 T 2 = p 3 V 3 T 3 76 × 45 ( 273 + 31 ) = p 2 × l ( 273 + 0 ) = p 3 ( 90 - l ) 273 + 273 76 × 45 304 = p 2 × l 273 = p 3 ( 90 - l ) 546 p 2 × l 273 = p 3 ( 90 - l ) 546 (Mercury column is at rest, so pressure difference p 2 - p 3 = 0 ⇒ p 2 = p 3 ) ∴ p 2 × l 273 = p 2 ( 90 - l ) 546 ⇒ 2 l = 90 - l ⇒ l = 30 cm 76 × 45 304 = p 2 × 30 273 ⇒ p 2 = 76 × 45 × 273 30 × 304 p 2 = 102.4