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1a)A 0.003525-mol sample of an organic compound was burned in oxygen in a bomb c

ID: 724687 • Letter: 1

Question

1a)A 0.003525-mol sample of an organic compound was burned in oxygen in a bomb calorimeter. The temperature of the calorimeter increased from 24.5oC to 26.7oC. If the heat capacity of the calorimeter is 4.46 kJ (oC)-1, then what is the constant volume heat of combustion of this compound, in kilojoules per mole? (Remember to include a "+" or "-" sign in your answer, as appropriate.)

Enter your answer with 3 significant figures

1b)What is w when a gas is compressed from 38.3 L to 28.8 L using a constant external pressure of 745 mmHg? (Remember to include a "+" or "-" sign as appropriate.)

Enter your answer in acceptable SI units, accurate to 3 significant figures. Use an acceptable SI abbreviation for units!

1c)What is w when 2.47 kg of H2O(l), initially at 25.0oC, is converted into water vapour at 137oC against a constant external pressure of 1.00 atm? Assume that the vapour behaves ideally and that the density of liquid water is 1.00 g/mL. (Remember to include a "+" or "-" sign as appropriate.)

Enter your answer in acceptable SI units, accurate to 3 significant figures. Use an acceptable SI abbreviation for units!

Explanation / Answer

1a)q= C*T, -q_sample=q_calorimeter

H= -q/# of moles

H= -C*T/ # of moles

H= -(4.46 kJ/°C)(26.7-24.5)/ .003525 mol

H= -2783= -2.78*10^3 kJ/mol

1b) w= -P*V

P= (745 mmHg)(1atm/760mmHg)= 377/380 atm

w= -(377/380)*(28.8-38.3)= 9.425 atm*L

w=1 atm * L = 101.32500 joules (taken from google)

w= (9.425 atm*L)(101.325 J/ atm*L)= 954.988= +955J

1c) Calculate the inital volume by using mass and density

V_i= (2.47kg)(1000g/kg)(1mL/1g)= 2470 mL= 2.47L

Use the Ideal gas law to calculate the final volume
PV=nRT
V= nRT/P
n= moles of water
n= (2.47kg)(1000g/kg)(1mole H2O/18.016g)= 137.1003552 mol H2O


V_f= (137.1003552)(.08206)(137+273.15)/1atm
V_f= 4614.37419L
w= -P*V

w= -P(V_f-V_i)

w= - (1atm)(4614.37419-2.47)

w= -4611.90419 atm*L

w= (-4611.90419 atm*L)(101.325 J/atm*L)= -467301J

w= -4.67*10^5 J

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