Calculate ΔU when 1 mole of water goes from 25°C to 1 atm and 30°C to 1 atm.
Answers
Answer:
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Explanation:
Volume of 0.5 mole of steam at 1 atm pressure
Volume of 0.5 mole of steam at 1 atm pressure=
Volume of 0.5 mole of steam at 1 atm pressure= P
Volume of 0.5 mole of steam at 1 atm pressure= PnRT
Volume of 0.5 mole of steam at 1 atm pressure= PnRT
Volume of 0.5 mole of steam at 1 atm pressure= PnRT =
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.0
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3L
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3L
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange=1×15.3=15.3litre−atm
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange=1×15.3=15.3litre−atm=15.3×101.3J=1549.89J
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange=1×15.3=15.3litre−atm=15.3×101.3J=1549.89Jw should be negative as the work has been done by the system on the surroundings
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange=1×15.3=15.3litre−atm=15.3×101.3J=1549.89Jw should be negative as the work has been done by the system on the surroundingsw=−1549.89J
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange=1×15.3=15.3litre−atm=15.3×101.3J=1549.89Jw should be negative as the work has been done by the system on the surroundingsw=−1549.89JHeat required to convert 0.5 mole of water at 100
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange=1×15.3=15.3litre−atm=15.3×101.3J=1549.89Jw should be negative as the work has been done by the system on the surroundingsw=−1549.89JHeat required to convert 0.5 mole of water at 100 o
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange=1×15.3=15.3litre−atm=15.3×101.3J=1549.89Jw should be negative as the work has been done by the system on the surroundingsw=−1549.89JHeat required to convert 0.5 mole of water at 100 o C to steam
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange=1×15.3=15.3litre−atm=15.3×101.3J=1549.89Jw should be negative as the work has been done by the system on the surroundingsw=−1549.89JHeat required to convert 0.5 mole of water at 100 o C to steam=0.5×40670J=20335J
Volume of 0.5 mole of steam at 1 atm pressure= PnRT = 1.00.5×0.0821×373 =15.3LChange in volume = Vol. of steam − Vol. of water=15.3− negligible =15.3LWork done by the system,w=P ext ×volumechange=1×15.3=15.3litre−atm=15.3×101.3J=1549.89Jw should be negative as the work has been done by the system on the surroundingsw=−1549.89JHeat required to convert 0.5 mole of water at 100 o C to steam=0.5×40670J=20335JAccording to the first law of