A cylinder of ideal gas is closed by a 4 kg movable piston of area 30 cm 2 . When the gas is heated from 50ºC to 100ºC, the piston raises 20 cm. The piston is then held in place and the gas is cooled back to 50ºC. If Q 1 is the heat added to the gas in the heating process and Q 2 is the heat lost during cooling, then (Q 1 – Q 2 ) can be approximated as 10 λ J, where λ is an integer between (0 to 9). Find λ .
Given: P atm = 1 × 10 5 N/m 2 g = 9.8 m/sec
2 
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(7)
During heating P = 
+ 1 × 10 5 = 1.13 × 10
5 N/m 2 Q 1 = Δ U 1 + Δ W 1 = Δ U 1 + P Δ V = Δ U 1 + 68J
During cooling Δ W = 0
Q 2
' = Δ U 2 = Δ U 1 (Heat supplied)
∴ Heat lost
Q 2 = –Q 2 ' = Δ U 1
∴ Q 1 – Q 2 = ( Δ U 1 + 68) – ( Δ U 1 ) = 68J ≈ ≈ 70 J
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