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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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An ideal gas has a molar heat capacity pressure $C_p = 2.5 R$. The gas is kept in a closed volume of $0.083 m^3$ at a temperature of $300 ^{\circ}K$ and a pressure of $10^5 N/m^2$. $200J$ of heat is supplied to the gas. Calculate the final temperature and pressure of the gas.

(A) $696 ^{\circ}K$, $1.02 \times 10^5 N/m^2$ (B) $69.6 ^{\circ}K$, $10.2 \times 10^5 N/m^2$ (C) $609.6 ^{\circ}K$, $1.02 \times 10^5 N/m^2$ (D) $6.96 ^{\circ}K$, $10.2 \times 10^5 N/m^2$
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$C_v = C_p - R = 1.5$
The amount of gas is given by $n = \large\frac{PV}{RT} $$ = \large\frac{10^5 \times 0.083}{8.3 \times 600} $$= 1.67$
Therefore, $\Delta u = n C_v \Delta T = \Delta Q$ (Because, work done = 0)
$\Rightarrow 200 = \large\frac{5}{3} $$\times 1.5R \times \Delta T \rightarrow \Delta T = 9.64K$
$\Rightarrow$ New Temperature $= 609.64 ^{\circ} K$
$\Rightarrow $ New Pressure $= \large\frac{609.64}{600}$$\times 10^5 = 1.02 \times 10^5$ $N/m^2$
answered Mar 11, 2014 by balaji.thirumalai
 

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