Discussion Forum



1)

An office room contains about 2000 moles of air. The change in the internal energy of this much air when it is cooled from $34^{0}C$ to $24^{0}C$ at a constant pressure of 1.0 atm is 

[ Use $\gamma_{air}$=1.4 and universal gas constant =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:

Option C

Explanation:

 Given, 

 Number  of moles of air in room (n)=2000= $2 \times 10^{3}$

 Temperature difference (dT)=24-34=$-10^{0} C$

 We know that,

 $dQ=nC_{v}dT= 2 \times 10^{3} \times \frac{R}{0.4}[-10]$

 = $\frac{-2 \times 10^{3} \times 8.314 \times 10}{0.4}$

  $=\frac{2 \times 8.314 \times 10^{5}}{4}= \frac{16.628}{4} \times  10^{5}$

   $=-4.2 \times 10^{5}$ J