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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 $\mathrm{5 atm}$ . During this process, it releases $\mathrm{800 J}$ of heat to the surroundings. Calculate the change in enthalpy of the gas
Option: 1
$\mathrm{9332.5 J}$

Option: 2
$\mathrm{564.40 J}$

Option: 3
$\mathrm{245.20 J}$

Option: 4
$\mathrm{687.35 J}$

Answers (1)

best_answer

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

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

Given:

$\mathrm{ Q =-800 \mathrm{~J} \quad }$ (negative because heat is released)

$\mathrm{ P =5 \mathrm{~atm} \quad }$ (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:

$\mathrm{ \Delta H=-800 \mathrm{~J}-(5 \mathrm{~atm}) \cdot\left(-0.02 \mathrm{~m}^3\right) \cdot 101325 \mathrm{~Pa} / \mathrm{atm} }$

$\mathrm{ \Delta H=-800 \mathrm{~J}+10132.5 \mathrm{~J} }$

$\mathrm{ \Delta H=9332.5 \mathrm{~J} }$

Therefore, the correct option is 1.

Posted by

himanshu.meshram

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