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Questions 4-12 · Q4

Q.A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system.

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Step 1. The gas expands against a constant external pressure p=1 atm=1.013×105p = 1\text{ atm} = 1.013\times10^5 Pa, from Vi=5V_i = 5 L to Vf=10V_f = 10 L, so ΔV=5\Delta V = 5 L =5×10−3= 5\times10^{-3} m³.

Step 2. Work done by the gas: W=p ΔV=(1.013×105 Pa)(5×10−3 m3)=506.5W = p\,\Delta V = (1.013\times10^5\text{ Pa})(5\times10^{-3}\text{ m}^3) = 506.5 J.

Step 3. Heat absorbed: Q=400Q = 400 J.

Step 4. By the First Law, ΔU=Q−W=400−506.5=−106.5\Delta U = Q - W = 400 - 506.5 = -106.5 J.

Step 5. This matches the textbook's printed answer of −106.5 J: the gas's internal energy decreases slightly, since it does slightly more work expanding than the heat it absorbed.

✓Final answer

ΔU = −106.5 J

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