A balloon filled with He initially has a volume of 876 L at 0.0degreeC and 1.0 a

Question

A balloon filled with He initially has a volume of 876 L at 0.0degreeC and 1.0 atm, then the temperature of the balloon is increased to 35degreeC while pressure remains constant. How many moles of helium are in the balloon? What is the final volume of the balloon? Calculate q, w. deltaU, and delta H for this process. Useful hints that can be applied in many cases and should be remembered: C_v,m = 3R/2 for any ideal monatomic gas. For ANY ideal gas, C_p,m - C_v,m = R. One mole of a perfect monatomic gas (C_v = 3/2 R) initially at 300 K and 12.0 atm is expanded to a final pressure of 1.0 atm. Calculate q, w, delta U, and deltaH if the process is (a) isothermal and reversible, and (b) adiabatic and reversible (T_f= 102 K). When three moles of oxygen gas (O_2) is heated at a constant pressure of 3.75 atm its temperature increases from 255K to 305K. Given that the molar heat capacity of O_2 at constant pressure is 29.4 J/(mol.K), calculate deltaV, w, q, deltaU, and deltaH. Calculate the change in enthalpy for the following reactions and indicate whether the reaction is endothermic or exothermic. Calculate the standard enthalpy of formation of N_2O_5 from the following data:

Explanation / Answer

First Use Charles' law to calculate the final volume as follows:

V1/T1=V2/T2

V1= 876 L, T1= 0.0C or 273 K and T2= 35+273=308K

V2 = V1/T1*T2

V2= 876 L, / 273 K *308K

V2= 988 L

Now calculate the number of moles of He as follows:

PV= nRT

n=PV/RT

n = 1.00 atm * 988 L / 0.08206 L atm / K.mol*308 K

n= 39.1 mol

molar heat capacity of helium = 20.8 J/degree Celsius x mol.

DeltaE = q + w
q = heat supplied = amount x specific heat x temperature change.
w = work done to system = - (work done by system) = - P DeltaV

Therefore,
q = (39.1 mole)*(20.8 J/°C*mol)*(35°C) = 28464.8J = 28.5 kJ
w = -(1 atm)* (988-876) = -112 L-atm

But 1 L-atm = 101.3 J, so
w = -112*101.3 = -11345.5 J = -11.3 kJ

Depta U = q + w = 28.5 – 11.3= 17.2 kJ

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