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Under constant pressure, the temperature of 2.43 mol of an ideal monatomic gas is raised 11.9 K. What are (a) the work W done by the gas, (b) the energy transferred as heat Q, (c) the change ΔEint in the internal energy of the gas, and (d) the change ΔK in the average kinetic energy per atom

1 Answer

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Answer:

Step-by-step explanation:

Given

no of moles
n=2.43

Temperature raised
\Delta T=11.9 k

Work done by gas


W=\int_(V_1)^(V_2)PdV


W=P\Delta V


W=nR\Delta T


W=2.43* 8.314* 11.9


W=240.41 kJ

(b)Energy Transferred as heat


Q=nc_p\Delta T


c_p=specific heat at constant Pressure


c_p for ideal Mono atomic gas is
(5R)/(2)


Q=2.43* (5R)/(2)* 11.9


Q=601.03 kJ

(c)Change in Internal Energy


\Delta U=Q-W


\Delta U=601.03-240.41=360.62 kJ

(d)Change in average kinetic Energy
\Delta k


K.E._(avg)=(3)/(2) * k* T


\Delta K.E.=(3)/(2) * k* \Delta T ,where k=boltzmann constant


\Delta K.E.=(3)/(2)* 1.38* 10^(-23)* 11.9


\Delta K.E.=2.46* 10^(-22) J

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