Answer:
A) if the system is isothermal then all the heat added to the system will be used to do work (since none is used to raise the temperature of the gas). The heat added will be equal to the work done = 340 J
B) change in internal energy of the system of the process is isothermal will be zero, since there is no rise in temperature.
C) an adiabatic process is one involving no heat loss or gain through the system, Therefore heat gain will be zero
D) if the process is adiabatic then there is no heat loss or gain through the system and hence there is no change in temperature. Change in internal energy will be zero
E) if the process is isobaric then, there is no work done and the total heat to the system is equal zero
F) if there is no work done, and no heat added, then the internal energy will be equal zero.
Answer:
is there an answier choice u have to choose from or no
Explanation:
Answer:
q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.
Explanation:
1. Heat absorbed
q = nΔH = 1 mol × (40.57 kJ/1 mol) = 40.57 kJ
2. Change in volume
V(water) = 0.018 L
pV = nRT
1 atm × V = 1 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 373.15 K
V = 30.62 L
ΔV = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L
3. Work done
w = -pΔV = - 1 atm × 30.60 L = -30.60 L·atm
w = -30.60 L·atm × (101.325 J/1 L·atm) = -3100 J = -3.10 kJ
4. Why the difference?
Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.
The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.
