Please answer 7,8 and 9 One mole of oxygen gas is at a pressure of 5.65 atm and
ID: 1532329 • Letter: P
Question
Please answer 7,8 and 9 One mole of oxygen gas is at a pressure of 5.65 atm and a temperature of 32.0 degree C. (a) If the gas is heated constant volume until the pressure triples, what is the final temperature? degree C (b) If the gas is heated so that both the pressure and volume are doubled, what is the final temperature? degree C (a) An ideal gas occupies a volume of 2.8 cm^3 at 20 degree C and atmospheric pressure. Determine the number of molecules of gas in the container. molecules (b) If the pressure of the 2.8-cm^3 at 2.0 times 10^-11 Pa (an extremely good vacuum) while the temperature remains constant, how many moles of gas remain in the container? mol What is the average kinetic energy of a molecule oxygen at a temperature of 260 K?Explanation / Answer
7) given
n = 1 mole
P1 = 5.65 atm = 5.65*1.013*10^5 pa
T1 = 32 C = 32 + 273 = 305 k
a) At constant volume, P/T = constant
P1/T1 = P2/T2
==> T2 = (P2/P1)*T1
= (3*P1/P1)*305
= 915 k
= 915 - 273
= 642 C
b) here, use, P1*V1/T1 = P2*V2/T2
==> T2 = (P2/P1)*(V2/V1)*T1
= (2*P1/P1)*(2*V1/V1)*T1
= 4*305
= 1220 k
= 1220 - 273
= 947 C
8)
a) let n is the number of moles of the gas.
n = P*V/(R*T)
= 1.013*10^5*2.8*10^-6/(8.314*(20+273)
= 1.16*10^-4 moles
no of molecules, N = n*Na
= 1.16*10^-4*6.023*10^23
= 6.97*10^19
b) n = P*V/(R*T)
= 2*10^-11*2.8*10^-6/(8.314*(20+273)
= 2.30*10^-20 mol
9) Average kinetic energy of an Oxygen molecule,
< KE > = (5/2)*k*T
= (5/2)*1.38*10^-23*260
= 8.97*10^-21 J
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