2 moles of a monoatomic gas are expanded to d | vedclass

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(B) Given: $n = 2$, $V_f = 2V_i$, $T_i = 300\, K$, process $P/V = k$ (polytropic index $x = -1$).Using $PV = nRT$ and $P = kV$, we get $kV^2 = nRT$, s

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$\Delta Q = 3200\, R$

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(B) Given: $n = 2$, $V_f = 2V_i$, $T_i = 300\, K$, process $P/V = k$ (polytropic index $x = -1$).Using $PV = nRT$ and $P = kV$, we get $kV^2 = nRT$, so $T \propto V^2$.$T_f = T_i (V_f/V_i)^2 = 300 	imes (2)^2 = 1200\, K$.Change in temperature $\Delta T = T_f - T_i = 1200 - 300 = 900\, K$. (Option $A$ is true).Work done $W = \int P dV = \int_V^{2V} kV dV = k [V^2/2]_V^{2V} = k/2 (4V^2 - V^2) = 3/2 kV^2 = 3/2 PV = 3/2 nRT$.At initial state, $W_i = 3/2 nRT_i = 3/2 	imes 2 	imes R 	imes 300 = 900\, R$. (Option $D$ is true).Change in internal energy $\Delta U = n C_v \Delta T = 2 	imes (3/2 R) 	imes 900 = 2700\, R$.Heat supplied $\Delta Q = \Delta U + W = 2700\, R + 900\, R = 3600\, R$. (Option $C$ is true).Since $\Delta Q = 3600\, R$, option $B$ is false.

$2$ moles of a monoatomic gas are expanded to double its initial volume, through a process $P/V = \text{constant}$. If its initial temperature is $300\, K$, then which of the following is not true?

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