Suppose 3.00 mol of an ideal gas of volume V_1 = 3.50 m^3 at T_1 = 300 K is allo

Question

Suppose 3.00 mol of an ideal gas of volume V_1 = 3.50 m^3 at T_1 = 300 K is allowed to expand isothermally to V_2 = 7.80 m^3 at T_2 = 300 K. (a) Determine the work done by the gas. (b) Determine the heat added to the gas. c) Determine the change in internal energy of the gas. In an engine, an almost ideal gas is compressed adiabatically to one-third its volume. In doing so, 2350 J of work is done on the gas. (a) How much heat flows into or out of the gas? (b) What is the change in internal energy of the gas? (c) Does its temperature rise or fall? rise fall

Explanation / Answer

5. (a) W = - n*R*T *ln( V/V ) where W = work done, n = moles, R = gas constant
= - 3* 8.314* 300K*ln( 7.8m³ / 3.5m³ )
= -5996.26 J

Internal energy of an ideal gas U = n*Cv*T
So the change in internal energy of isothermal process always zero:
(c) U = 0

On the other hand change of internal energy equals heat transferred to the gas plus work done on the gas:
U = Q + W
Where Q is heat and W is work
Therefore
(b) Q = U - W = 0J - (-5996.26) = 5996.26J

6. (a) In an adiabatic process: Q=0.

(b) U = QW = 0(2350) = 2350 J.

(c) U= (3 / 2)* nRT so with an increase in internal energy increases the temperature. so rise is the correct option.

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