The initial state of a certain gas is $(P_i, V_i, T_i)$. It undergoes expansion until its volume becomes $V_f$. Consider the following two cases:
$(a)$ The expansion takes place at constant temperature.
$(b)$ The expansion takes place at constant pressure.
Plot the $P-V$ diagram for each case. In which of the two cases is the work done by the gas more?
$(a)$ The expansion takes place at constant temperature.
$(b)$ The expansion takes place at constant pressure.
Plot the $P-V$ diagram for each case. In which of the two cases is the work done by the gas more?
(B) The situation is shown in the provided $P-V$ graph,where the variation is shown for each process.
Process $1$ represents the isobaric expansion (constant pressure),and process $2$ represents the isothermal expansion (constant temperature).
Since the work done by a gas during expansion is equal to the area under the $P-V$ curve,we compare the areas under the two curves.
From the graph,the area under the curve for process $1$ (isobaric) is clearly larger than the area under the curve for process $2$ (isothermal).
Therefore,the work done by the gas is more in the isobaric process (case $b$).
Process $1$ represents the isobaric expansion (constant pressure),and process $2$ represents the isothermal expansion (constant temperature).
Since the work done by a gas during expansion is equal to the area under the $P-V$ curve,we compare the areas under the two curves.
From the graph,the area under the curve for process $1$ (isobaric) is clearly larger than the area under the curve for process $2$ (isothermal).
Therefore,the work done by the gas is more in the isobaric process (case $b$).
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