Five moles of Hydrogen gas initially at STP i | vedclass

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(C) Given: Number of moles $n = 5$, initial temperature $T_1 = 273 \, K$ (at $STP$), final temperature $T_2 = 673 \, K$, gas constant $R = 8.3 \, J \,

Vedclass · Accepted Answer

$41.50$

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(C) Given: Number of moles $n = 5$, initial temperature $T_1 = 273 \, K$ (at $STP$), final temperature $T_2 = 673 \, K$, gas constant $R = 8.3 \, J \, mol^{-1} \, K^{-1}$, and adiabatic index $\gamma = 1.4$.For an ideal gas, the change in internal energy $\Delta U$ is given by the formula: $\Delta U = n C_v \Delta T$.Since $C_v = \frac{R}{\gamma - 1}$, we substitute this into the equation:$\Delta U = n \left( \frac{R}{\gamma - 1} ight) (T_2 - T_1)$.Substituting the given values:$\Delta U = 5 	imes \left( \frac{8.3}{1.4 - 1} ight) 	imes (673 - 273)$.$\Delta U = 5 	imes \left( \frac{8.3}{0.4} ight) 	imes 400$.$\Delta U = 5 	imes 8.3 	imes 1000$.$\Delta U = 41500 \, J = 41.5 \, kJ$.Thus, the increase in internal energy is $41.5 \, kJ$.

Five moles of Hydrogen gas initially at $STP$ is compressed adiabatically so that its temperature becomes $673 \, K$. The increase in internal energy of the gas is $(R=8.3 \, J \, mol^{-1} \, K^{-1}, \gamma=1.4$ for diatomic gas$)$ (in $kJ$)

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