1 mole of rigid diatomic gas performs a work | vedclass

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(B) From the first law of thermodynamics,$Q = \Delta U + W.$Given that $W = Q/5,$ we have $Q = \Delta U + Q/5.$Therefore,$\Delta U = Q - Q/5 = 4Q/5.$F

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) From the first law of thermodynamics,$Q = \Delta U + W.$Given that $W = Q/5,$ we have $Q = \Delta U + Q/5.$Therefore,$\Delta U = Q - Q/5 = 4Q/5.$For a rigid diatomic gas,the internal energy change is $\Delta U = n C_v \Delta T,$ where $C_v = \frac{5}{2}R.$Also,$Q = n C \Delta T,$ where $C$ is the molar heat capacity.Substituting these into the equation $\Delta U = \frac{4}{5}Q,$ we get $n C_v \Delta T = \frac{4}{5} (n C \Delta T).$$C_v = \frac{4}{5} C \implies C = \frac{5}{4} C_v.$Substituting $C_v = \frac{5}{2}R,$ we get $C = \frac{5}{4} 	imes \frac{5}{2} R = \frac{25}{8} R.$Comparing this with $\frac{xR}{8},$ we find $x = 25.$

$A$. Steam engine	$I$. Magnetic confinement of plasma
$B$. Electron microscope	$II$. Laws of thermodynamics
$C$. Non-reflecting coatings	$III$. Wave nature of electrons
$D$. Tokamak	$IV$. Interference of light

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

$1$ mole of rigid diatomic gas performs a work of $Q/5$ when heat $Q$ is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{xR}{8}.$ The value of $x$ is $\ldots \ldots \ldots .$ $[R =$ universal gas constant $]$

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