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  • Calculation of Work Done and En- thalpy Change A gas expands isothermally from a volume of 10 L to 20 L against a constant external pressure of 2 atm. calculate the enthalpy change (H) fo

Calculation of Work Done and En- thalpy Change

A gas expands isothermally from a volume of 10 L to 20 L against a constant external pressure of 2 atm. calculate the enthalpy change (H) for the process.

Option: 1

2026 kJ
 


Option: 2

2300 J
 


Option: 3

2500 J
 


Option: 4

2900 J


Answers (1)

best_answer

The work done (w) during an isothermal expansion of a gas against a constant
external pressure can be calculated using the equation:

\mathrm{ w=-P_{\mathrm{ext}} \Delta V }

where \mathrm{ P_{\mathrm{ext}} } is the external pressure and ?V is the change in volume.
Given:

\mathrm{ \begin{aligned} & P_{\text {ext }}=2 \text { atm } \\ & \Delta V=20 \mathrm{~L}-10 \mathrm{~L} \end{aligned} }

Substituting the values and calculating the work done (w):

\mathrm{ w=-(2 \mathrm{~atm}) \times(20 \mathrm{~L}-10 \mathrm{~L}) \times 101.325 \mathrm{~J} / \mathrm{L} \cdot \mathrm{atm} }

The enthalpy change (?H) for an isothermal process can be calculated using
the equation:

\mathrm{ \Delta H=w }

since \mathrm{ \Delta U }(change in internal energy) for an isothermal process is zero. Therefore, the work done (w) by the gas is:

\mathrm{ w=-(2 \mathrm{~atm}) \times(20 \mathrm{~L}-10 \mathrm{~L}) \times 101.325 \mathrm{~J} / \mathrm{L} \cdot \mathrm{atm} }

And the enthalpy change (?H) for the process is:

\mathrm{ \begin{aligned} &\Delta H=-w\\ &\Delta H=2026 k J \end{aligned} }

 

 

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Pankaj

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