n = 3.66 mol of Hydrogen gas is initially at T = 325 K temperature and p i = 2.7

Question

n = 3.66 mol of Hydrogen gas is initially at T = 325 K temperature and pi = 2.74×105 Pa pressure. The gas is then reversibly and isothermally compressed until its pressure reaches pf = 6.62×105 Pa. What is the volume of the gas at the end of the compression process? Answer: 1.49E-2 m^3

b. How much work did the external force perform? Answer: 8.72E3 J

c. How much heat did the gas emit?

d. How much entropy did the gas emit?

e. What would be the temperature of the gas, if the gas was allowed to adiabatically expand back to its original pressure?

Explanation / Answer

a) Initialy

moles = 3.66 mole , Temperature = 325 K Pressure = 2.74×105 Pa

so volume = nRT / P (ideal gas equation) = 3.66 X R X 325 / 2.74×105 Pa

Final pressure = 6.62×105 Pa

Vf = 3.66 X 8.314 X 325 / ( 6.62××105) m^3 = 1.49X 10^-2 m^3

b) For an isothermal process, the work is

W = nRT ln(Vi/Vf)
Since PiVi = PfVf, we have Vi/Vf = Pf/Pi and

W = 3.66*8.31*324 ln(6.62 * 10^5/ (2.74*10^5)) J

W = 8698.6 J

c) S = nCv ln(Tf/Ti) + nR ln(Vf/Vi) for the entropy change of an ideal gas

For an isothermal process, Tf = Ti and

S = nR ln(Vf/Vi)

From PiVi = PfVf, we have Vf/Vi = Pi/Pf so

S = nR ln(Pi/Pf) = 3.66*8.31 ln(2.74 * 10^5/ (6.62*10^5)) J/°K

S = -26.84 J/°K <=== is the change of entropy

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