A thin U-tube sealed at one end consists of three bends of length ℓ=250mm each, forming right angles. The vertical parts of the tube are filled with mercury to half the height.All of mercury can be displaced from the tube by heating slowly the gas in the sealed end of the tube, which is separated from the atmospheric air by mercury. Determine the work done (in joule) by the gas there by if the atmospheric pressure is P0=105Pa, the density of mercury is ρmer=13.6×103 kg/m3, and the cross-sectional area of the tube is S=1 cm2

The work done by the gas is the sum of work done W1 against the force of atmospheric pressure and the work done W2 against the gravity. Thus total work done,W=W1+W2The mercury-gas interface is shifted upon the complete displacement of mercurys=2ℓ+ℓ/2=5ℓ2 and hence W1=Fs =P0S×5ℓ2=5P0Sℓ2The work done W2 against the gravity is equal to the change in the potential energy of mercury as a result of its displacement. The whole of mercury rises as a result of displacement by ℓ relative to the horizontal part of the tube. This quantity regarded as the final height of the center of gravity of the whole mercury. The initial position of center of gravity of mercury is =l/8.Thus  W2=U2−U1​=Mg(ℓ−ℓ/8)=78Mgℓ​ where  M=2ℓSρmer​∴W=W1+W2​=52P0Sℓ+74ρmer gSℓ2​=7.74J