Question

1. a) Suppose 2 moles of a monatomic ideal gas occupies 5 m³ at a pressure of 1600 Pa.

First: Find the temperature of the gas in Kelvin.

Answer: 481 K

Second: Find the total internal energy of the gas.

Answer:12000 J

1. b) Suppose the gas undergoes an isobaric expansion to a volume of 7 m³.

(Don’t forget to include + and – in each of the problems below)

First: Find Q

Answer: 8000 J

Second: Find W

Answer: -3200 J

Third: Find ?U

Answer: 4800 J

1. c) Suppose the gas then undergoes adiabatic expansion to a pressure of 900 Pa and a volume of 11.5 m³.

First: Find Q

Answer: 0 K

Second: Find W

Answer: -1275 J

Third: Find ?U

Answer: -1275 J

1. d) The gas then undergoes isothermal compression to a volume of 5 m³.

First: Find Q

Answer: -8620.6 J

Second: Find W

Answer: 8620.6 J

Third: Find ?U

Answer: 0 J

1. e) The gas then undergoes an isochoric process and returns to 1600 Pa of pressure.

First: Find Q

Answer: -3525 J

Second: Find W

Answer: 0 J

Third: Find ?U

Answer: -3525 J

1. f) Find the change in internal energy for the entire PV cycle.

Answer: 0 J

Please help me figure out how to get all the answers

Answer 1

1a)T=PV/nR=(1600*5)/(2*8.314)=481K

1second)Internal Energy=1.5*n*R*T=12000 J

1b)for isobaric ,Tnew=673.4K

W=P*change in volume=-3200J

chnage in internal energy=nCv*change in temperature=-4800J

Q=U-W=8000 J

1c)for adiabtic

Expression for work done is given by

for monoatomic=5/3=1.67

Q=0

W=-1275 J

U=-1275 J

1d)for isothermal

U=0

W=nRT*ln(V2/V1)=8620.6 J

Q=-8620.6J

1e)for isochoric

W=0 (as volume is unchanged)

U=Q=nCv*change in tempeature=-3525 J

1f) For a closed cycle (if initial and final points are same),internal energy change is 0

so 0