For a particular process that is carried out at constant pressure, q = 140 kJ and w = -30 kJ. Therefore,
For a particular process that is carried out at constant pressure, = 140 kJ and = -30 kJ. Therefore,
ΔE = 110 kJ and ΔH = 140 kJ.
ΔE = 140 kJ and ΔH = 110 kJ.
ΔE = 140 kJ and ΔH = 170 kJ.
ΔE = 170 kJ and ΔH = 140 kJ.
How do I approach this problem? thanks

Question

For a particular process that is carried out at constant pressure, q = 140 kJ and w = -30 kJ. Therefore,
For a particular process that is carried out at constant pressure,  = 140 kJ and  = -30 kJ. Therefore,
	 ΔE = 110 kJ and ΔH = 140 kJ.
	 ΔE = 140 kJ and ΔH = 110 kJ.
	 ΔE = 140 kJ and ΔH = 170 kJ.
	 ΔE = 170 kJ and ΔH = 140 kJ.
How do I approach this problem? thanks

aaronq · Answer

At constant P, $\Delta H=q\; and \; \Delta E=q+w$