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  • A gas expands from a volume of to <img alt="\mathrm{2.0 \mathrm

A gas expands from a volume of \mathrm{1.0 \mathrm{~L}} to \mathrm{2.0 \mathrm{~L}} at a constant pressure of \mathrm{1.0 \mathrm{~atm}} and absorbs \mathrm{50 \mathrm{~J}} of heat from the surroundings during the expansion. Calculate the change in internal energy of the gas.

Option: 1

-151.3 J


Option: 2

151.3 J


Option: 3

-400 J


Option: 4

202 J


Answers (1)

best_answer

The first law of thermodynamics states that:

                                     \mathrm{ \Delta U=q+w }

where \mathrm{\Delta U} is the change in internal energy of the system, q is the heat added to or removed from the system, and w is the work done by or on the system.

Since the pressure is constant, we can use the following equation to calculate the work done by or on the gas:
                                  \mathrm{ w=-P \Delta V }

where P is the pressure and \mathrm{\Delta V} is the change in volume.
Substituting the values given, we get:

                   \mathrm{ w=-(1.0 \mathrm{~atm})(2.0 \mathrm{~L}-1.0 \mathrm{~L})=-1.0 \mathrm{~L} \mathrm{~atm} }

Since \mathrm{1 \mathrm{~L} \, \, atm =101.3 \mathrm{~J},} we can convert this to joules:

                 \mathrm{ w=(-1.0 \mathrm{~L} \mathrm{~atm})(101.3 \mathrm{~J} / \mathrm{L} \text { atm })=-101.3 \mathrm{~J} }

Substituting this value and \mathrm{q=-50 \mathrm{~J}} into our equation for \mathrm{\Delta U}, we get:

                    \mathrm{ \Delta U=q+w=(-50 \mathrm{~J})+(-101.3 \mathrm{~J})=-151.3 \mathrm{~J} }
Therefore, the change in internal energy is -151.3 So, correct option is (A)

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Ritika Harsh

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