Three moles of a monatomic ideal gas are heated at a constant volume of 2.90 m3. The amount of heat added is 5.10 103 J.
(a) What is the change in the temperature of the gas?
(b) Find the change in its internal energy.
(c) Determine the change in pressure.
(a) ΔU = Q + W Because volume does not change, no work is done on gas: W = - ∫ p dV = 0 Hence: ΔU = Q The change of internal energy of an ideal gas is: ΔU = n·Cv·ΔT The heat capacity of an ideal monatomic gas is Cv = (3/2)·R Therefore: n·(3/2)·R·ΔT = Q => ΔT = Q / (n·(3/2)·R) = 5.10×10³J / (3mol · (3/2) · 8.314472J/molK) = 136.31 K Temperature rises by 136.31K (b) as shown in (a) ΔU = Q = 5.10×10³J (c) use ideal gas law p·V = n·R·T => p = n·R·T/V => Δp = p₂ - p₁ = n·R·T₂/V - n·R·T₁/V = n·R·ΔT/V = 3mol · 8.314472J/molK · 136.31K / 2.90 m³ = 1 172.426 Pa Pressure rises by 1172.426Pa
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