A tank with a constant volume of 1.45 m 3 contains 15.2 moles of a monatomic ide

Question

A tank with a constant volume of 1.45 m3 contains 15.2 moles of a monatomic ideal gas. The gas is initially at a temperature of 300 K. An electric heater is used to transfer 57400 J of energy into the gas. It may help you to recall thatCV = 12.47 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.

What is the temperature of the gas after the energy is added?

What is the change in pressure of the gas?

How much work was done by the gas during this process?

Explanation / Answer

Solution: Constant volume of the container (hence the gas) V = 1.45 m3

Number of moles of the monoatomic ideal gas, n = 15.2

Initial temperature of the gas, Ti = 300 K

The specific heat at constant volume of the gas, Cv = 12.47 J/K.mole

Electric heater transfers Q = 57400 J of heat energy to the gas.

Part (A)

For the isochoric process (constant volume process), we have

Q = n*Cv*(Tf – Ti)

where Tf is the final temperature of the gas,

57400 J = (15.2 mole)*(12.47 J/K.mole)* (Tf – Ti)

(Tf – Ti) = 302.83 K

Tf = 302.83 K + Ti

Tf = 302.83 K + 300 K

Tf = 602.83 K

Thus the final temperature of the gas is 602.83 K when the energy is added.

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Part (B)

From the ideal gas equation P*V = n*R*T, we can write,

P*V = n*R*T

P = n*R*T/V

we notice that n, R and Vin the above equation are constants as they do not change when process takes place (n and R are constants and we kept V fixed thus net term is a constant number).

P = (15.2 mole)(8.31 J/K.mole)*(302.83 K)/(1.45 m3)

P = 26380 N/m2

P = 26380 Pa

Thus the change in pressure is 26830 N/m2

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part (3) Since the volume is constant, change in the volume, V = 0,

W = P*V = 0 J

Thus the work done by the gas during this process is 0.00 J

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