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(C) The change in internal energy $(\Delta U)$ of an ideal gas is given by the formula $\Delta U = n C_v \Delta T$.Given:Number of moles $(n) = 2000 =

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$-4.2 	imes 10^5 	ext{ J}$

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(C) The change in internal energy $(\Delta U)$ of an ideal gas is given by the formula $\Delta U = n C_v \Delta T$.Given:Number of moles $(n) = 2000 = 2 	imes 10^3 	ext{ mol}$.Initial temperature $(T_i) = 34^{\circ} C$, Final temperature $(T_f) = 24^{\circ} C$.Temperature change $(\Delta T) = 24 - 34 = -10 	ext{ K}$.For air, $\gamma = 1.4$. The molar heat capacity at constant volume is $C_v = \frac{R}{\gamma - 1} = \frac{R}{1.4 - 1} = \frac{R}{0.4}$.Substituting the values:$\Delta U = n \left( \frac{R}{0.4} ight) \Delta T$$\Delta U = (2 	imes 10^3) 	imes \left( \frac{8.314}{0.4} ight) 	imes (-10)$$\Delta U = - \frac{2 	imes 8.314 	imes 10^4}{0.4}$$\Delta U = - \frac{16.628 	imes 10^4}{0.4} = -41.57 	imes 10^4 	ext{ J} \approx -4.2 	imes 10^5 	ext{ J}$.

List-$I$	List-$II$
$A$. Isothermal Process	$I$. Work done by the gas decreases internal energy
$B$. Adiabatic Process	$II$. No change in internal energy
$C$. Isochoric Process	$III$. The heat absorbed goes partly to increase internal energy and partly to do work
$D$. Isobaric Process	$IV$. No work is done on or by the gas

Curve	Process
$I$	$a)$ Adiabatic
$II$	$b)$ Isobaric
$III$	$c)$ Isochoric
$IV$	$d)$ Isothermal

An office room contains about $2000$ moles of air. The change in the internal energy of this much air when it is cooled from $34^{\circ} C$ to $24^{\circ} C$ at a constant pressure of $1.0 \text{ atm}$ is (Use $\gamma_{\text{air}} = 1.4$ and universal gas constant $R = 8.314 \text{ J/mol-K}$)

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What is the relationship between the slopes of isothermal and adiabatic curves?

For the given $P-V$ diagram of a thermodynamic system,match the curves with their respective thermodynamic processes. ($P$ = Pressure and $V$ = Volume)
Curve Process
$I$ $a)$ Adiabatic
$II$ $b)$ Isobaric
$III$ $c)$ Isochoric
$IV$ $d)$ Isothermal

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