Question

Three moles of an ideal monatomic gas expand at a constant pressure of 2.90atm : the volume of the gas changes from 3.30*10^-2m^3 to 4.50*10^-2m^3.

Part A, Calculate the initial temperature of the gas.

Part B, Calculate the final temperature of the gas.

Part C, Calculate the amount of work the gas does in expanding.

Part D, Calculate the amount of heat added to the gas.

Part E, Calculate the change in internal energy of the gas.

Answer #1

A)

P = pressure = 2.90 atm = 2.90 x 1.01 x 10^{5} Pa = 2.94
x 10^{5} Pa

V_{i} = initial volume = 0.033 m^{3}

T_{i} = initial temperature = ?

n = number of moles = 3

Using the equation

PV_{i} = nRT_{i}

(2.94 x 10^{5}) (0.033) = (3) (8.314) T_{i}

T_{i} = 389 K

b)

P = pressure = 2.90 atm = 2.90 x 1.01 x 10^{5} Pa = 2.94
x 10^{5} Pa

V_{f} = final volume = 0.045 m^{3}

T_{f} = initial temperature = ?

n = number of moles = 3

Using the equation

PV_{f} = nRT_{f}

(2.94 x 10^{5}) (0.045) = (3) (8.314) T_{f}

T_{f} = 530.4 K

C)

work done is given as

W = P (V_{f} - V_{i}) = (2.94 x 10^{5})
((0.045) - (0.033)) = 3528 J

d)

Q = n C_{p} (T_{f} - T_{i}) = (3) (2.5)
(8.314) (530.4 - 389) = 8817 J

e)

Using first law

Q = W + U

8817 = 3528 + U

U = 5289 J

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