5 moles of Hydrogen left(gamma=frac{7}{5}righ | vedclass

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(B) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = n C_v \Delta T$.For an adiabatic process,the work done $W$ is eq

Vedclass · Accepted Answer

$41.55$

Vedclass · Answer

(B) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = n C_v \Delta T$.For an adiabatic process,the work done $W$ is equal to the negative change in internal energy,but the question asks for the increase in internal energy,which is $\Delta U = n C_v (T_2 - T_1)$.Given:$n = 5 \ mol$$T_1 = 0^{\circ} C = 273.15 \ K$$T_2 = 400^{\circ} C = 673.15 \ K$$\Delta T = T_2 - T_1 = 400 \ K$$\gamma = 7/5 = 1.4$$C_v = \frac{R}{\gamma - 1} = \frac{8.30}{1.4 - 1} = \frac{8.30}{0.4} = 20.75 \ J \ mol^{-1} \ K^{-1}$.Calculating $\Delta U$:$\Delta U = 5 	imes 20.75 	imes 400 \ J$$\Delta U = 5 	imes 20.75 	imes 0.4 \ kJ$$\Delta U = 41.50 \ kJ$.Rounding to the nearest provided option,the increase in internal energy is $41.55 \ kJ$.

$5$ moles of Hydrogen $\left(\gamma=\frac{7}{5}\right)$ initially at $S.T.P.$ are compressed adiabatically so that its temperature becomes $400^{\circ} C$. The increase in the internal energy of the gas in kilo-joules is $(R=8.30 \ J \ mol^{-1} \ K^{-1})$.

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