Answer to Question #123934 in Physical Chemistry for Troma Aaronmaid Kollie

Question

Calculate ∆H, w, q and ∆U for heating 1.00 mol of argon from 100 K to 300 K at a constant volume of 30.6 L.

Answer 1

In isochoric process, the work (w) done is zero (since dW = PdV = 0 when V = constant).

∆Q = c_v m(T₂ - T₁)

c_v for Ar is 0.312 kJ/kg*K

1 mole of Ar is 39.95 g or 0.03995 kg

T₂ - T₁ = 300 - 100 = 200K

∆Q = 0.312 x 0.03995 x 200 = 2.49 kJ

The first law of thermodynamics: dU = dQ – dW. In this equation dW is equal to dW = pdV. Then: dU = dQ – pdV. For isochoric process (pdV = 0): dU = dQ . ∆U = ∆Q = 2.49 kJ

∆H = ∆U + P∆V. As P∆V = 0, Thus in the isochoric process heat supplied to the system is used for increasing the internal energy of the system. ∆H = ∆U = 2.49 kJ.