A total of 48 ,J heat is given to one mole of | vedclass

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(D) According to the $1^{	ext{st}}$ law of thermodynamics:$\Delta Q = \Delta U + W$Here,$\Delta Q = 48 \,J$,$n = 1 \,mol$,and $\Delta T = 2 \,K$ (sin

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$23.1$

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(D) According to the $1^{	ext{st}}$ law of thermodynamics:$\Delta Q = \Delta U + W$Here,$\Delta Q = 48 \,J$,$n = 1 \,mol$,and $\Delta T = 2 \,K$ (since a change of $2^{\circ}C$ is equal to a change of $2 \,K$).For a monatomic gas like helium,the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.The change in internal energy is $\Delta U = n C_v \Delta T = (1) \left(\frac{3}{2} Right) (2) = 3R$.Substituting the values into the first law equation:$48 = 3R + W$$W = 48 - 3(8.3)$$W = 48 - 24.9$$W = 23.1 \,J$.

$A$ total of $48 \,J$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^{\circ} C$. The work done by the gas is : (Given,$R=8.3 \,J \,K^{-1} \,mol^{-1}$.) (in $\,J$)

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