An ideal monatomic gas undergoes an adiabatic compression from state 1 with pres

Question

An ideal monatomic gas undergoes an adiabatic compression from state 1 with pressure p1=1 atm, volume V1=8 L, and temperature T1=300 K to state 2 with pressure p2=32 atm, volume V2=1 L.
(a) What is the temperature of the gas in state 2?
(b) How many moles of gas are present?

(c) What is the average translational kinetic energy per mole before and after the compression?

(d) What is the ratio of the squares of the rms speeds before and after the compression?

(e) If we do not know that the ideal gas here is monatomic, demonstrate that the gas is truly monatomic.

Explanation / Answer

b)state 1 : PV = nRT

1*8 = n 300*0.082

n = 0.325 moles

a) state 2 : PV =nRT

32 * 1 = 0.325 * 0.082 T

T = 1200 k

c)translational energy = 3/2 PV

KE before = 12atm l = 1224.49 J

KE after = 48atm l = 4897.96J

d)v= rms speed

1/2 mv2 = 1224.49 j

v before = 2448.98/m

v after = 9795.92/m

v before : v after

2448.98 :9795.92

1:4

e) for adiobatic process P V is constant

if we substitute the PV values in two states and equate them we get =5/3

hence we can prove that the given gas is monoatomic

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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