A sample of gas occupies a volume of 57.4 mL. As it expands, it does 140.5 J of

Question

A sample of gas occupies a volume of 57.4 mL. As it expands, it does 140.5 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas? How much work must be done on a system to decrease its volume from 11.0L + 4.0L by exerting a constant pressure of 3.0 atm? When 1651 J of heat energy is added to 43.7 g of hexane C_6H_14, the temperature increase by 16.7.C. Calculate the molar hear capacity of C_6H_14. Liquid sodium is being considered as on engine coolont. How many grams of liquid sodium (minimum) are needed to absorb 4.80 MS of energy (in form of heat) if the temperature of the sodium is not to increase by more than 10.0 degree C? Use C_p 30.8 J (k middot mol) for No(1) at 500 k. The following equation is the balanced combustion reaction for C_6H_6. 2C_6(l) + 15O_2 (g) rightarrow 12 CO_2(g)+6H_2O(e) + 6542kj IF 9.000 g of C_6H_6 is burned and the heat produced from the burning is added to 4691_g of water at 21 degree C, what is the Final temperature of the water? For a particular isomer of C_8H_18, the following reaction produces 5113.3 KJ of heat per mole of C_8H_18 (g) consumed user standard conditions. What is the standard enthalpy of Formation of this isomer of C_8H_18(g)?

Explanation / Answer

1.(a): Since pressure is constant, this is an example of isobaric process. In isobaric process, the work done is calculated as

W = P x dV = P x (Vf - Vi) ------ (1)

Givene P = 783 torr = 104391 Pa(N/m2)

W = 140.5 J

Vi = 57.4 mL = 57.4 mL x (1 m3 / 106 mL) = 5.74x10-5 m3

Vf = ?

Now from eqn(1)

W = 140.5 J = 104391 Pa x (Vf - 5.74x10-5 m3)

=> Vf = 1.4033 x 10-3 m3 =  1.4033 x 10-3 m3 x (106 mL / m3) = 1403 mL (answer)

(b):

Since pressure is constant, this is an example of isobaric process. In isobaric process, the work done is calculated as

W = P x dV = P x (Vf - Vi) ------ (1)

Givene P = 3.0 atm = 303975 Pa(N/m2)

dV = Vf - Vi = 11.0 L - 4.0 L = 7.0 L = 7.0 L x (1 m3 / 1000L) = 7.0x10-3 m3

Now from eqn(1)

W = PxdV = 303975 Pa x 7.0x10-3 m3 = 2128 J = 2.13 kJ (answer)

(c): Given the mass of hexane = 43.7 g

Molecular mass of hexane = 86.18 g/mol

Hence moles of hexane in the container = mass / molar mass = 43.7 g / 86.18 g/mol = 0.5071 mol

Q = n x c x dT

=> 1651 J = 0.5071 mol x c x 16.7 DegC

=> c = 1651 J / (0.5071 mol x 16.7 DegC ) = 195 J/molxDegC (answer)

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