One mole of an ideal gas performs 900 J of w | vedclass

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(D) According to the first law of thermodynamics,$\Delta U = q + w$. Since the gas performs work on the surroundings,$w = -900 \ J$. The internal ener

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$1525$

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(D) According to the first law of thermodynamics,$\Delta U = q + w$. Since the gas performs work on the surroundings,$w = -900 \ J$. The internal energy increases,so $\Delta U = +625 \ J$. Substituting these values: $625 = q - 900$,which gives $q = 1525 \ J$. For an ideal gas,the enthalpy change is defined as $\Delta H = \Delta U + P\Delta V$. Since $w = -P\Delta V$,we have $P\Delta V = -w = 900 \ J$. Therefore,$\Delta H = 625 \ J + 900 \ J = 1525 \ J$.

One mole of an ideal gas performs $900 \ J$ of work on the surroundings. If the internal energy increases by $625 \ J$,find the value of $\Delta H$. (in $J$)

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