a) The molar volume of Me2X3 at 25.000°C and 1.0000 atm is 32.000cm3/mol. The CT

Question

a) The molar volume of Me2X3 at 25.000°C and 1.0000 atm is 32.000cm3/mol. The CTE, , is 25.000 x 10-6 K-1 and isothermal compressibility, , is 300.000 x 10-7 atm-1. Assume that and are constant for this range of pressures and temperatures. Calculate the molar volume at 375.000°C and 10,000. atm to five significant digits. Note that, as Me2X3 is not expected to be a perfect gas, PVnRT b)Three moles of a perfect monatomic gas are initially contained at 300.0°C and 202.0 kPa. When 37.0 kJ of heat are transferred to the gas, it reversibly expands and does 3100. J of work against its surroundings. What is the final temperature of the gas (K)?

Explanation / Answer

a)

Me2X3 does not follow ideal gas law. Hence the real gas law followed is

RT = ( P + /Vm2 ) (Vm - )

where R = gas constant = 82.057 Cm3 atm K-1 mol-1

T = temperature in K = (375+273.15) = 648.15 K

P = Pressure in atm = 10000 atm

= 25*10-6 K-1

= 300*10-7 atm-1

Vm = molar volume in Cm3 mol-1

Putting values in equation and solving we get

RT = ( P + /Vm2 ) (Vm - )

82.057 * 648.15 = (10000 + 25*10-6/Vm2) (Vm - 300*10-7)

53185.24455 = (10000 + 25*10-6/Vm2) (Vm - 300*10-7)

Vm = 5.3185544 Cm3/mol

b)

For given monoatomic gas

n = no. of moles = 3

T1 = Initial Temperature = 3000C = 573.15 K

P1 = Initial Pressure = 202.0 kPa

Heat Transferred (Q) = +37 kJ (since heat is transferred to the system)

Work done (W)= -3100 J (The work is done on the system)

Gas constant (R) = 8.314 J/K mol

The gas is reversibly expands

By the first law of thermodynamics

Q = U + W

37000 = U - 3100

40100 J = U

We know U = 3/2 nRT

40100 = 3/2 * 3 * 8.314 * T

T = 1071.8199 K

T = T2 - T1

1071.8199 = T2 - 573.15

T2 = 1644.96 K = 1371.820C (Final temperature)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.