Question

Please solve the following problems. You must show all work.

1. A 10.0 cm radius piston compresses a gas isothermally from a height of 15.0 cm to 2.50 cm at a constant pressure of 2.0 atm.

a) How much heat was added to the gas?

b) Now if 7000 J of heat is added to the system and the piston is only moves 5.0 cm up, what is the change in the internal energy of the system is the pressure is again constant at 2.0 atm?

2. Sketch a PV diagram for the following process:

a) A 2.0 L gas undergoes an isovolumetric increase in pressure from 1.0 atm to 2.0 atm

b) An isothermal compression from 2.0 atm and 2.0 L to 1.0 atm and 1.0 L

c) An isobaric compression from 2.0 L to 1.0 L

3. An ideal gas expands at a constant total pressure of 2.5 atm from 3.45 L to 6.70 L. Heat then flows out of the gas at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. Calculate:

a) the total work done by the gas in the process

b) the total heat flow into the gas.

4. Heat flows into an ideal gas at a constant volume. The pressure increases from 1.5 atm to 5.5 atm. Next the gas is compressed at constant pressure from 5.0 L to 2.5 L and goes back to its original temperature.

a) What is the total work done on the gas in the process?

b) What is the total change in internal energy?

c) What is the total heat flow of the process?

Answer 1

1)

Thermodynamic law : Q = ?U + W
W = PdV = 2*(3.14* 0.1^2 * 0.15 - 3.14*0.1* 0.025)
W =331.25 J
a. dQ = dU + dW = 331.25 J
b) dW = 2* 3.14*0.1*0.1* 0.05 = 0.00314 = = 318.82 J
dQ = 7000 + 318.82 = 7318.82 J

(or)

For an ideal gas undergoing an isothermal process (temperature remains constant) then the CHANGE in the internal energy is zero
dQ=dU +dW
dQ=0 +dW
dQ=dW

Work Done = pdV
dQ=pdV
pdV= (2.0 x 1.01 x 10^5) x pi x 0.1^2 x (0.15 - 0.025)
pdV=793 J

dQ=793 J

2)

3)