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Question: An office room contains about \({{2000}}\)moles of air. The change in the internal energy of this mu...

An office room contains about 2000{{2000}}moles of air. The change in the internal energy of this much air when it is cooled from 340Cto240C{{3}}{{{4}}^{{0}}}{{C to 2}}{{{4}}^{{0}}}{{C}}at a constant pressure of 1.0{{1}}{{.0}} atm is
(Use γair=1.4{{{\gamma }}_{{{air}}}}{{ = 1}}{{.4}}at universal gas constant=8,314J/molK{{ = 8,314 J/mol K}})
A. 1.9×105J{- 1.9 \times 1}{0}^{5}J
B. +1.9×105J{+ 1.9 \times 1}{0}^{5}J
C. 4.2×105J{- 4.2 \times 1}{0}^{5}J
D. +0.7×105J{+ 0.7 \times 1}{0}^{5}J

Explanation

Solution

First of all, find the temperature difference by subtracting the initial temperature from initial temperature and then substitute this value in the formula for change in internal energy, dQ=nCVdT{{dQ = n}}{{{C}}_{{V}}}{{dT}} where n={{n = }} number of moles, CV={{{C}}_{{V}}}{{ = }} heat capacity at constant volume and dT={{dT = }} temperature difference and then evaluate.
Use formula for heat capacity at constant volume, CV=Rγ{{{C}}_{{V}}}{{ = }}\dfrac{{{R}}}{{{\gamma }}} and substitute this value of CV{{{C}}_{{V}}} in above formula and find out the change in internal energy.

Complete step by step answer:
Given: Number of molecules of air in room, n=2000{{n = 2000}}moles
Initial temperature, Ti=340C{{{T}}_{{i}}}{{ = 3}}{{{4}}^{{0}}}{{ C}}
Final temperature, Tf=240C{{{T}}_{{f}}}{{ = 2}}{{{4}}^{{0}}}{{ C}}
Temperature difference is given by
dT=TfTi dT=240C340C=100C  {{dT = }}{{{T}}_{{f}}}{{ - }}{{{T}}_{{i}}} \\\ {{dT = 2}}{{{4}}^{{0}}}{{C - 3}}{{{4}}^{{0}}}{{C = - 1}}{{{0}}^{{0}}}{{C}} \\\
Pressure, P=1.0atm{{P = 1}}{{.0 atm}}(constant)
γair=1.4{{{\gamma }}_{{{air}}}}{{ = 1}}{{.4}}
Universal gas constant, R=8,314J/molK{{R = 8,314 J/mol K}}
Internal energy is defined as the total amount of kinetic energy (K.E.) and potential energy (P.E.) of all the particles in the system. When energy is given to increase the temperature then particles speed up and gain kinetic energy (K.E.).
Formula for change in internal energy at constant pressure is given by
dQ=nCVdT{{dQ = n}}{{{C}}_{{V}}}{{dT}}
Where n={{n = }} number of moles
CV={{{C}}_{{V}}}{{ = }} Heat capacity at constant volume
dT={{dT = }} Temperature difference
On substituting the values in above formula, we get
dQ=2000×R1.4×(10) dQ=2000×8.3141.4×(10)=4.2×105J  {{dQ = 2000 \times }}\dfrac{{{R}}}{{{{1}}{{.4}}}}{{ \times ( - 10)}} \\\ \Rightarrow {{dQ = 2000 \times }}\dfrac{{{{8}}{{.314}}}}{{{{1}}{{.4}}}}{{ \times ( - 10) = - 4}}{{.2 \times 1}}{{{0}}^{{5}}}{{ J}} \\\
Thus, the change in internal energy of the air is dQ=4.2×105J{{dQ = - 4}}{{.2 \times 1}}{{{0}}^{{5}}}{{ J}}.

Hence, the correct answer is option (C).

Note: In thermodynamics, the ratio of specific heat capacity at constant pressure and specific heat capacity at constant volume is an adiabatic index represented by γ{{\gamma }}. It is very important point to note that Cp{{{C}}_{{p}}} is always greater than CV{C_{{V}}} because at constant pressure, heat is absorbed for increasing internal energy and for doing work. But at constant volume, heat is absorbed only for increasing the internal energy and not for doing any work on the system.