__Enthalpy Calculator:__

__Enthalpy Calculator:__

Enter the values of internal energy at initial time, Q_{1(kJ)} and final time, Q_{2(kJ)}, pressure, p_{(k.Pa)} and change in volume at initial time, V_{1(m3)} and final time, V_{2(m3)} to determine the value of Enthalpy, H_{(kJ)}.

__Enthalpy Formula:__

__Enthalpy Formula:__

Enthalpy is a thermodynamic property of a system, representing the total heat content. Enthalpy is useful in describing energy changes in processes occurring at constant pressure, such as chemical reactions and phase changes. It is typically measured in joules (J) or kilojoules (kJ).

Change in enthalpy – represents the net energy transfer between the system and its surroundings at constant pressure. A positive H indicates the system gained energy, while negative H signifies energy loss.

**Heat Transfer (Q):** This term accounts for the thermal energy exchanged between the system and its surroundings. If the system absorbs heat (Q positive), the enthalpy change (H) will be positive. Conversely, if the system releases heat (Q negative), the enthalpy change will be negative.

**Pressure and Volume Change (p * (V _{2} – V_{1})):** This term considers the work done by the system or the surroundings due to changes in pressure and volume.

- If the system expands against a constant external pressure (V
_{2}> V_{1}), the system does work on the surroundings, and this term is positive. This can contribute to a positive or negative H depending on the heat transfer (Q). - If the system is compressed (V
_{2}< V_{1}), work is done on the system by the surroundings, and this term is negative.

Enthalpy, H_{(kJ)} in kilojoules is calculated by taking difference in internal energy at initial time, Q_{1(kJ)} and final time, Q_{2(kJ)} in kilojoules then add pressure, p_{(k.Pa)} in kilo Pascals and multiplied by change in volume at initial time, V_{1(m3)} and final time, V_{2(m3)} in cubic metres.

Enthalpy, H_{(kJ)} = (Q_{2(kJ)} – Q_{1(kJ)}) + p_{(k.Pa)} * (V_{2(m3)} – V_{1(m3)})

H_{(kJ)} = enthalpy in kilojoules, kJ.

Q_{1(kJ)} = internal energy at initial time in kilojoules, kJ.

Q_{2(kJ)} = internal energy at final time in kilojoules, kJ.

V_{1(m3)} = volume at initial time in cubic metres, m^{3}.

V_{2(m3)} = volume at final time in cubic metres, m^{3}.

p_{(k.Pa)} = pressure in kilo Pascals, k.Pa.

__Enthalpy Calculation:__

__Enthalpy Calculation:__

**Finding Change in Enthalpy using Heat Quantities and Volume Changes****:**

Given: Q_{1(kJ)} = 100kJ, Q_{2(kJ)} = 200kJ, V_{1(m3)} = 1m^{3}, V_{2(m3)} = 1.5m^{3}, p_{(k.Pa)} = 101.3k.Pa.

Enthalpy, H_{(kJ)} = (Q_{2(kJ)} – Q_{1(kJ)}) + p_{(k.Pa)} * (V_{2(m3)} – V_{1(m3)})

H_{(kJ)} = (200 – 100) + 101.3 * (1.5 – 1)

H_{(kJ)} = 100 + 101.3 * 0.5

H_{(kJ)} = 150.65kJ.

**Finding Final Heat Quantity using Change in Enthalpy, Pressure, and Volume Changes****:**

Given: Q_{1(kJ)} = 50kJ, H_{(kJ)} = 250kJ, V_{1(m3)} = 0.5m^{3}, V_{2(m3)} = 1m^{3}, p_{(k.Pa)} = 101.3k.Pa.

Enthalpy, H_{(kJ)} = (Q_{2(kJ)} – Q_{1(kJ)}) + p_{(k.Pa)} * (V_{2(m3)} – V_{1(m3)})

250 = (Q_{2(kJ)} – 50) + 101.3 * (1 – 0.5)

250 = (Q_{2(kJ)} – 50) + 101.3 * 0.5

250 = Q_{2(kJ)} – 50 + 50.65

250 = Q_{2(kJ)} – 0.65

Q_{2(kJ)} = 250 – 0.65

Q_{2(kJ)} = 249.35kJ.

__Enthalpy Calculations:__

__Enthalpy Calculations:__