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  • Question : Enthalpy Change and Pressure-Volume Work A gas is compressed from an initial volume of <img alt="\mathrm{0.04\; m^{3}}" src="https://entrancecorner.oncodecogs.

Question : Enthalpy Change and Pressure-Volume Work

A gas is compressed from an initial volume of \mathrm{0.04\; m^{3}} to a final volume of \mathrm{0.02\; m^{3}} at a constant pressure of 5 atm. During this process, it releases 800 J of heat to the surroundings. Calculate the change in enthalpy of the gas.

Option: 1

9332.5 J


Option: 2

564.40 J


Option: 3

245.20 J


Option: 4

687.35 J


Answers (1)

The change in enthalpy\mathrm{(\Delta H)} is given by:

\mathrm{\Delta H=Q-P \Delta V}

Given:
Q = −800 J (negative because heat is released)
P = 5 atm (pressure)

\mathrm{\Delta V=V_f-V_i=0.02 \mathrm{~m}^3-0.04 \mathrm{~m}^3=-0.02 \mathrm{~m}^3} (negative due to compression)

Substitute the values into the formula:

\mathrm{\begin{aligned} & \Delta H=-800 \mathrm{~J}-(5 \text { atm }) \cdot\left(-0.02 \mathrm{~m}^3\right) \cdot 101325 \mathrm{~Pa} / \mathrm{atm} \\ & \Delta H=-800 \mathrm{~J}+10132.5 \mathrm{~J} \\ & \Delta H=9332.5 \mathrm{~J} \end{aligned}}

Therefore, the correct option is A.

Posted by

Ramraj Saini

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