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 $1.0\, atm$ is
[Use $\gamma_air = 1.4$ and Universal gas cons $\tan \, t = 8.314\, J / mol \, K $ ]

• 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 $(d T)=24-34=-10^{\circ} C$
We know that,
$d Q =n C_{v} d T=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 $