A cylinder with a piston contains 0.200 mol of nitrogen at 1.70×10 5 Pa and 320

Question

A cylinder with a piston contains 0.200 mol of nitrogen at 1.70×105 Pa and 320 K . The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure.

Part A

Find the work done by the gas during the initial compression.

-266

Correct

Part B

Find the heat added to the gas during the initial compression.

Correct

Part C

Find internal-energy change of the gas during the initial compression.

Correct

Part D

Find the work done by the gas during the adiabatic expansion.

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Part E

Find the heat added to the gas during the adiabatic expansion.

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Part F

Find the internal-energy change of the gas during the adiabatic expansion.

-266

Explanation / Answer

(D)

Use the following formula,

       PV = nRT

(1.7x 105 pa) V = (0.2) (8.31) (320)

                      V = 0.00313 m3

During compression,

V1= initial volume = 0.00313 m3

V2= final volume = V1/2 = 0.00313/2 m3 = 0.00156 m3

T1 = initial temperature = 320 K

Use the formula,

V1/T1 = V2/T2

0.00313 /320 = 1.56 x 10-3 /T2

                 T2 = 150 K

During adiabatic expansion,

P1 = initial pressure = 1.7 x 105 pa

V1 = initial Volume = 1.56 x 10-3

V2= final volume = 0.00313

P2 = final pressure

For adiabatic process

P1 V1 = P2 V2

For diatomic gas nitrogen, = 1.4

(1.7 x 105 ) (1.56 x 10-3 )1.4 = P2 (0.00313)1.4

                                                     P2 = 6.44 x 104 pa

The work done is given as,

W = (P2 V2 - P1 V1) / (1 - )

W = ((6.44 x 104 pa ) (0.00313) - (1.7 x 105 ) (1.56 x 10-3 )) / (1-1.4)

W = 158.93 J

(E)

Heat added = Q =0

(F)

From the first law of thermodynamics,

Q = U + W

U = Q - W = 0 – 158.93 J

U = -158.93 J

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