When an ideal gas expands isothermally from 5 | vedclass

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(B) The pressure $P = 10^7 \ N \cdot m^{-2}$.Initial volume $V_i = 5 \ m^3$.Final volume $V_f = 10 \ m^3$.Change in volume $\Delta V = V_f - V_i = 10

Vedclass · Accepted Answer

$-50$

Vedclass · Answer

(B) The pressure $P = 10^7 \ N \cdot m^{-2}$.Initial volume $V_i = 5 \ m^3$.Final volume $V_f = 10 \ m^3$.Change in volume $\Delta V = V_f - V_i = 10 \ m^3 - 5 \ m^3 = 5 \ m^3$.The work done on the gas during expansion against constant external pressure is given by $W = -P_{ext} \Delta V$.Substituting the values: $W = -(10^7 \ N \cdot m^{-2}) 	imes (5 \ m^3) = -50 	imes 10^6 \ J$.Since $1 \ MJ = 10^6 \ J$,we get $W = -50 \ MJ$.

When an ideal gas expands isothermally from $5 \ m^3$ to $10 \ m^3$ at $25 \ ^{\circ}C$ against a constant pressure of $10^7 \ N \cdot m^{-2}$,then the work done on the gas is (in $MJ$)

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