Question

Two moles of an ideal gas is changed from its initial state ($16atm,6L$) to final state ($4atm,15L$) in such a way that this change can be represented by a straight line in $P-V$ curve. The maximum temperature attained by the gas during the above change is:

(Take $R=\frac{1}{12}Latm{K}^{-1}{mol}^{-1}$)

• A. $324K$
• B. $604K$
• C. $1296K$
• D. $9782K$

Answer 1

Answer: (B)

$604K$

Given $P_1=16$ atm

$P_2=4$ atm

$V_1=6$ L

$V_2=15$ L

The shaded portion of the above graph represents the work done

Area=area of the triangle= $ABC$+Area of rectagle$BCED$

=$\frac{1}{2}\times 9\times 12+9\times 4=90LAtm=W$

W$=2.303\times nRTlog\frac{V_2}{V_1}$-

$90=2.303\times 2\times 0.083\times T\times \log \frac{15}{6}$

$T=604$ K