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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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An ideal monoatomic gas at temperature $27^{\circ}C$ and pressure $10^6N/m^2$ cal of heat is added to the system without changing the volume. Calculate the final temperature of the gas. $(R = 8.31 \: J/mol-K \: and\: J = 4.18 \: J/cal)$

$\begin {array} {1 1} (A)\;860^{\circ}C & \quad (B)\;833\: K \\ (C)\;560^{\circ}C & \quad (D)\;860 \: K \end {array}$

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For $n$ mole of gas, we have $PV = nRT$
Here, $P = 10^6\: N/m^2 , \: V = 10L = 10^{-2}m^3 \: and \: T = 27^{\circ}C = 300 \: K$
So, $n = \large\frac{PV}{RT+} = \large\frac{10^6 \times 10^{-2}}{(8.31 \times 300)}$$ = 4$
For ‘monoatomic’ gas, $C_V = \large\frac{3R}{2}$
Thus, $C_V = \large\frac{3 \times 8.31\: (J/mol-K) }{ 2}$$ = \large\frac{3 \times 8.31}{(2 \times 4.18)} $$ \sim 3\: cal/(mole-K)$
Let $ \Delta T$ be the rise in temperature when $n$ mole of gas is given $Q$ cal of heat at constant volume.
Then, $Q = nC_V \Delta T$
$ \Rightarrow \Delta T = \large\frac{Q}{nC_V}$$ = 10,000\: cal/[4\: mole \times 3\: cal(mole-K)] = 833\: K$
Thus, final temperature of gas is, $T + \Delta T = 300 + 833\: K = 1133\: K = 860^{\circ}C$
Ans : (A)
answered Mar 15, 2014 by thanvigandhi_1
edited Mar 9 by sharmaaparna1
 

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