Question

An office room contains about $2000$ moles of air. The change in the internal energy of this much air when it is cooled from $34^{\circ }C$ to $24^{\circ }C$ at a constant pressure of $1.0atm$ is
[Use ${\gamma }_{a}ir=1.4$ and Universal gas cons $\tan t=8.314J/molK$ ]

• A. $-1.9\times 10^{5}J$
• B. $+1.9\times 10^{5}J$
• C. $-4.2\times 10^{5}J$
• D. $+0.7\times 10^{5}J$

Answer 1

Answer: (C)

Given,
Number of moles of air in room $(n)=2000=2\times 10^3$
Temperature difference $(dT)=24-34=-10^{\circ }C$
We know that,
$dQ=n{C}_{v}dT=2\times 10^3\times \frac{R}{0.4}[-10]$
$=\frac{-2\times 10^3\times 8.314\times 10}{0.4}$
$=\frac{2\times 8.314\times 10^5}{4}=\frac{16.628}{4}\times 10^5$
$=-42\times 10^{5}J$