I have 0.5 mol xenongas in a metal (non-expanding) 2 L box at 25 °C. If I put it

Question

I have 0.5 mol xenongas in a metal (non-expanding) 2 L box at 25 °C. If I put itover a candle and add 1.0 kJ of energy, the gas is heated up to110 °C. Calculate q, w, U,and H for thisprocess.

Explanation / Answer

volume is constnat, so W = 0 U = Q + W = Q + 0 = Q U = 1.0 kJ recall the legendre transform, H = U + PV  => H = U + (PV) for this case, V is constant, so H = U + V*P P = Pfinal - Pinitial, both from VDW Equation of State Xenon     a = 4.250 L2*bar/mol2 = 4.19442388L2*atm/mol2 b = 0.05105 (L/mol) P = nRT/(V - nb) - an2/V2 n = 0.5 mol, Tinit = 298K,   Tfinal = 383 K, V = 2 L Pinit = (0.5 mol*0.0821 L*atm/mol-K * 298K)/(2 L-0.5 mol * 0.05105L/mol)-4.1944 L2*atm/mol2 *(0.5mol)2/(2L)2 Pinit = 5.93337033 atm Pfinal =(0.5 mol*0.0821 L*atm/mol-K * 383K)/(2L- 0.5 mol * 0.05105L/mol)-4.1944 L2*atm/mol2 *(0.5mol)2/(2L)2 Pfinal = 7.70054895 V*P = 2 L*(7.70054895 atm - 5.93337033 atm)*(1m3/1000L)* (101325 Pa / 1 atm) V*P = 358.118747 J = 0.358 kJ H = 1 kJ + 0.358 kJ = 1.358 kJ = 1.36 kJ

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