The following parts are unrelated to each other. (a) What is the change in heat
ID: 1022028 • Letter: T
Question
The following parts are unrelated to each other.
(a) What is the change in heat q during a reversible isothermal compression of 1.50 moles of an ideal gas at a fixed temperature T = 298 K from a volume of 350.0 L to 200.0 L?
(b) Suppose 3.50 moles of an ideal gas undergo a reversible adiabatic expansion in a piston from a temperature of 395 K, where the volume is 100.0 L, to a temperature of 298 K. What is the final volume necessary to cause this reduction in temperature? What is the change in internal energy of the gas as a result of this adiabatic expansion?
(c) Consider a chamber filled with gaseous N2 (molar mass 28.0 g/mol) and O2 (molar mass 32.0 g/mol) which behave ideally (cv = (3/2)R). The chamber has a volume of 1.00 L and is initially at a temperature T = 25.0 OC ; the total initial pressure of all
gases is ptot,i = 760. torr. The initial partial pressure of oxygen is pO2,i = 593. torr. A vacuum pump attached to the chamber is turned on. The total pressure inside the chamber is reduced to 35 mtorr (i.e. milli-torr) and the vacuum pump is turned off. You may assume that the temperature does not change as the pressure is reduced. What is the final partial pressure of O2: pO2,f?
Explanation / Answer
From 1st law, deltaU= Q+W
for reversible expansion under constant temperature, deltaU=0, work done =
Q= -W
W= nRT ln (P2/P1), n= number of moles, R= gas constant =8.314 J/mol.K T= 298K
W= nRT ln (P2/P1)= 1.5*8.314*298* ln (V1/V2)= 1.5*8.314* 298*ln(350/200) = 2080 Joules
Q= -W = -2080 joules
b) since the proces is adiabatic, Q=0
deltaU= W= nR(T2-T1)/ (Y-1)
for adiabatic process, T1V1(Y-1)= T2 V2(Y-1)
Y= ratio of specific heats
Assuming the gas to be diatomic, for which Y= 1.4
395* (100) 0.4 = 298*(V2) 0.4
V2= 202.3 L
delta U= nR(T2-T1)/ (Y-1) =3.5* 8.314*( 298-395)/ 0.4=-7057 Joules
c) This is the case of isothermal process
moles of gas initially, n =
PV/RT
P= 760 Torr =760/760 atm, V =1 L, R= 0.0821 L.atm/mole.K T= 25 deg.c =25+273.15= 298.15
n= PV/RT = 1*1/ (0.0821*298.15)= 0.041 moles
Partial pressure of oxygen= 593 Torr, partial pressure of nitrogen = 760-593= 167 mm Hg
moles of n2 = (593/760)*0.041 =0.032 moles, moles of oxygen =0.041-0.032 =0.009 moles
during the expansion, volume increases and number of moles rermain the same. Moles of gas will not change
Parital pressure of oxygen = (0.032/0.041)*35*10-3 Torr =27*10-3 Torr, partial pressure of nitrogen = 35*10-3-27*10-3= 8*10-3 Torr
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