Workshop 5: Heat and Work/ Enthalpy Thermodynamics is all about the study of ene

Question

Workshop 5: Heat and Work/ Enthalpy Thermodynamics is all about the study of energy transfers. The first law of states that AEq+w. This means that all energy changes can be computed work. In today's workshop you will compute energy changes for a gaseous system consider calorimetry in terms of heat and The first problem deals with an ideal gas. To solve this problem you need to remember som concepts about gases. Specifically you will need to remember the ideal gas equation: R-0.08206 L atm /mol K PV nRT. To compute the energy of a gaseous system: E (3/2) nRT (R 8.314]/mole-K). 1. Consider a system that contains 2 moles of an ideal gas in a cylindrical syringe. Initially the temperature of the system is T1-300 K and the pressure is 1 atm. Now, the gas is slowly heated (at a constant pressure of 1 atm) until the gas reaches a temperature of T2 350 K. During this constant pressure process, the volume expands to a higher value Vz and the energy of the gaseous system changes from Ei to E2. a) Compute the change in energy (AE- E2 E) for this gas in units of Joules. b) Find the change in volume (AV) for this gas in units of L.

Explanation / Answer

1.

(a) ?E = E2 - E1 = (3/2)nR?T

= 1.5×2×8.314×(350-300) = 1247.1 J

(b) change in volume(?V) = nR(?T)/P

= 2× 0.0821×(350-300)/1

= 8.21 L

(c) w = -p?V = -1×8.21×101 = - 829.21J

Work is doing by gas.yes negative sign means work is done by gas(system)

(d). ?E = q + w

q = 1247.1 - 829.21 = 417.89 J

Answer-2:

Heat loss by Al = Heat by water

mc(?T) = ms(?T)

50×0.88×(225-T) = 100×4.18×(T-20)

T = 39.5 ?

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