Question

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?

_____K

(b) Find the change in its internal energy.

_____J

(c) Determine the change in pressure.

_____Pa

Answer 1

(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