An ideal gas at 300 , K expands from a volume | vedclass

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Question

(A) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} 	imes \Delta V$. Here,$P_{ext} = 1 	i

Vedclass · Accepted Answer

$-900 \, J$

Vedclass · Answer

(A) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} 	imes \Delta V$. Here,$P_{ext} = 1 	imes 10^5 \, N \, m^{-2}$. Change in volume $\Delta V = V_f - V_i = (1 	imes 10^{-2} - 1 	imes 10^{-3}) \, m^3 = (10 	imes 10^{-3} - 1 	imes 10^{-3}) \, m^3 = 9 	imes 10^{-3} \, m^3$. Substituting the values: $W = -(1 	imes 10^5) 	imes (9 	imes 10^{-3}) = -900 \, J$. Since the gas expands,work is done by the system,hence the value is negative.

An ideal gas at $300 \, K$ expands from a volume of $1 \times 10^{-3} \, m^3$ to $1 \times 10^{-2} \, m^3$ against a constant external pressure of $1 \times 10^5 \, N \, m^{-2}$. Calculate the work done.

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