The work done by an ideal gas of 2 text{ mole | vedclass

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(C) The work done by an ideal gas during an isothermal process is given by the formula: $W = nRT \ln\left(\frac{V_f}{V_i}\right)$.For the first case:

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$3W$

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(C) The work done by an ideal gas during an isothermal process is given by the formula: $W = nRT \ln\left(\frac{V_f}{V_i}ight)$.For the first case: $n_1 = 2 	ext{ moles}$,$T_1 = T$,$V_i = V$,$V_f = 2V$.$W_1 = W = 2RT \ln\left(\frac{2V}{V}ight) = 2RT \ln 2$.For the second case: $n_2 = 4 	ext{ moles}$,$T_2 = \frac{T}{2}$,$V_i = V$,$V_f = 8V$.$W_2 = n_2 R T_2 \ln\left(\frac{V_f}{V_i}ight) = 4R \left(\frac{T}{2}ight) \ln\left(\frac{8V}{V}ight)$.$W_2 = 2RT \ln 8 = 2RT \ln(2^3) = 2RT \cdot 3 \ln 2$.Since $W = 2RT \ln 2$,we have $W_2 = 3W$.

The work done by an ideal gas of $2 \text{ moles}$ in increasing its volume from $V$ to $2V$ at constant temperature $T$ is $W$. The work done by an ideal gas of $4 \text{ moles}$ in increasing its volume from $V$ to $8V$ at constant temperature $\frac{T}{2}$ is:

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