The P – V diagram for 2 moles of an ideal gas undergoing a process A→B is shown in figure. The maximum temperature of the gas during the process is

Equation of Straight line AB isP=mV+c …(1)Where m→ slopec→ Intercept2 P 0 =m V 0 +c and P 0 =m 2 V 0 +c here, m=−P0V0and c=3P0 ∵PV=nRT⇒P=nRTV Putting P in equation (1)T= 1 nR m V 2 +cV …(2)T will be maximum when, dT dV =0& d 2 T d V 2 <0 Putting, dT dV =0 ⇒V=−c2m Tmax=−c24nRm=−14nR×3P02−P0/V0=9P0V04nR

The P – V diagram for 2 moles of an ideal gas undergoing a process A→B is shown in figure. The maximum temperature of the gas during the process is

Equation of Straight line AB isP=mV+c …(1)Where m→ slopec→ Intercept2 P 0 =m V 0 +c and P 0 =m 2 V 0 +c here, m=−P0V0and c=3P0 ∵PV=nRT⇒P=nRTV Putting P in equation (1)T= 1 nR m V 2 +cV …(2)T will be maximum when, dT dV =0& d 2 T d V 2 <0 Putting, dT dV =0 ⇒V=−c2m Tmax=−c24nRm=−14nR×3P02−P0/V0=9P0V04nR

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The P – V diagram for 2 moles of an ideal gas undergoing a process $A \to B$ is shown in figure. The maximum temperature of the gas during the process is

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3 P 0 V 0 2 R

2 P 0 V 0 R

3 P 0 V 0 4 R

9 P 0 V 0 8 R

answer is A.

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Detailed Solution

Equation of Straight line AB is
$P = m V + c$ …(1)
Where $m →$ slope
$c →$ Intercept
$2 P 0 = m V 0 + c$
and $P 0 = m 2 V 0 + c$
here, $m = \frac{- P_{0}}{V_{0}} a n d c = 3 P_{0}$
$∵ P V = n R T \Rightarrow P = \frac{n R T}{V}$
Putting P in equation (1)
$T = 1 n R m V 2 + c V$ …(2)
T will be maximum when, $d T d V = 0 & d 2 T d V 2 < 0$
Putting, $d T d V = 0$
$\Rightarrow V = - \frac{c}{2 m}$
$T_{\max} = - \frac{c^{2}}{4 n R m} = - \frac{1}{4 n R} \times \frac{{(3 P_{0})}^{2}}{- P_{0} / V_{0}} = \frac{9 P_{0} V_{0}}{4 n R}$