five grams of nitrogen gas at initial pressure of 3.0 atm and 20 degrees C under

Question

five grams of nitrogen gas at initial pressure of 3.0 atm and 20 degrees C undergo an isobaric expansion until the volume has tripled a) what is the gas volume after expansion? b) What is the gas temperature after expansion in(degrees C)? the gas pressure is then decreased at constant volume until the original temperature is reached. c) what is the gas pressure after the decrease? finally the gas is isothermally compressed until it returns to its initial volume d) what is the final gas pressure?

Explanation / Answer

a) The molar mass of N2 is 28 g/ mol
The number of moles is n = 5/28 = 0.1786 mol
The initial conditions are p1 =3 atm and T1 = 20 C = 20+273 = 293 k
Using Ideal Gas law , we find initial volume
V1 = nRT / p1
= 0.1786*8.31*293/(3*101325)
= 1.430 *10^-3 m^3
= 1430.58 cm^3
an isobaric expansion until the volume triples results in V2 = 3 V1
V2= 4291.74 cm^3 ---> Answer

b) After the expansion
p2V2/T2 = p1V1/T1
=> T2 = [p2V2/p1V1]*T1
= 1*3*T1 = 3T1
=3*293
T2= 879 K or 606 deg C ---> Answer

c) a constant volume decrease at V3 =V2 = 4291.74 cm^3 back to T3= T1= T2 results in the following
p3V3/T3 = p2V2/T2
=> p3 = [T3V2/T2V3]*p2
= (1/3)*1*p2
= (1/3)*1*3
p3= 1 atm ---> Answer

d)an isothermal compression at T4 =T3 back to initial volume V4= V1= V3 results in the following
p4V4/T4 = p3V3/T3
=> p4 = [T4V3/T3V4]*p3
= 3*1 atm
p4= 3 atm ---> Answer

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