The work done by 6 moles of helium gas when i | vedclass

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Question

(C) For an ideal gas undergoing a process at constant pressure,the work done $W$ is given by the formula $W = nR\Delta T$.Here,the number of moles $n

Vedclass · Accepted Answer

$997.2$

Vedclass · Answer

(C) For an ideal gas undergoing a process at constant pressure,the work done $W$ is given by the formula $W = nR\Delta T$.Here,the number of moles $n = 6$.The universal gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$.The change in temperature $\Delta T = 20^{\circ} C = 20 \ K$ (since the change in temperature is the same in Celsius and Kelvin scales).Substituting these values into the formula:$W = 6 	imes 8.31 	imes 20$$W = 120 	imes 8.31$$W = 997.2 \ J$.Therefore,the correct option is $C$.

The work done by $6$ moles of helium gas when its temperature increases by $20^{\circ} C$ at constant pressure is (Universal gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$) (in $J$)

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