Figure (b) represents a process involving 3.24 moles of a monatomic ideal gas ex

Question

Figure (b) represents a process involving 3.24 moles of a monatomic ideal gas expanding from 0.100 m3to 0.200 m3.

Find the work done on the system.
W =  J

Find the change in the internal energy of the system.
?U =  J

Find the thermal energy transferred in the process.
Q =  J

EXAMPLE 12.9A General Process GOAL Find thermodynamic quantities for a process that doesn't fall into any previously discussed category .00 P(10 Pa) P(10 Pa) .00 FA 2.00 2.00 A1 PROBLEM A quantity of 4.00 moles of a monatomic ideal gas expands from an initial volume of 0.100 m to a final volume of 0.300 m and pressure of 2.5 x 105 Pa (Figure (a)). Compute (a) the work done on the gas, (b) the change in internal energy of the gas, and transferred to the gas 1.00 A 1.00 A2 V(m*) V(m3) 0.100 0.200 0.300 0.100 0.200 0.300 (c) the thermal energy STRATEGY The work done on the gas is equal to the negative of the area under the curve in the PV diagram. Use the ideal gas law to get the temperature change and, subsequently, the change in internal energy. Finally, the first law gives the thermal energy transferred by heat.

Explanation / Answer

n = number of moles = 3.24

Pi = initial pressure = 3 x 105 Pa

Vi = initial Volume = 0.1 m3

Ti = initial temperature

Using formula

Pi Vi = n R Ti

(3 x 105) (0.1) = 3.24 x 8.314 Ti

Ti = 1113.7 K

Pf = final pressure = 1 x 105 Pa

Vf = initial Volume = 0.2 m3

Tf = initial temperature

Using formula

Pf Vf = n R Tf

(1 x 105) (0.2) = 3.24 x 8.314 Tf

Tf = 742.5 K

Change in Temperature = T = Ti - Tf= 1113.7 - 742.5 = 371.2 K

a)

Work done = Area A1 + Area A2

w = 0.5 (2 x 105) (0.2 - 0.1) + (0.1) (1 x 105) = 20,000 J

b)

U = (3/2) n R T

U = 1.5 x 3.24 x 8.314 (371.2)

U = -14998.72 J           since temperature is decreasing

c)

Using first law of thermodyanamics

Q = U + W

Q = -14998.72 + 20,000

Q = 5001.3 J

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