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(B) For an ideal gas,the equation of state is $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is mass,$M$ is molar mass).Thus,$PV = \frac{m}{M} RT$,which impl

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$B$

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(B) For an ideal gas,the equation of state is $PV = nRT$,where $n = \frac{m}{M}$ ($m$ is mass,$M$ is molar mass).Thus,$PV = \frac{m}{M} RT$,which implies $P = \left( \frac{mR}{MV} \right) T$.For the initial state,the slope of the line $A$ is $S_1 = \frac{mR}{MV}$.In the new state,the mass $m' = 2m$ and the volume $V' = \frac{V}{2}$.The new slope $S_2$ is given by $S_2 = \frac{m'R}{MV'} = \frac{(2m)R}{M(V/2)} = 4 \left( \frac{mR}{MV} \right) = 4S_1$.Since the slope $S_2$ is greater than $S_1$,the new line must be steeper than line $A$. Looking at the graph,line $B$ has a greater slope than line $A$. Therefore,the correct line is $B$.

The pressure versus temperature graph of an ideal gas at constant volume $V$ and mass $m$ is shown by the straight line $A$. If the mass of the gas is doubled and the volume is halved,then the corresponding pressure versus temperature graph will be shown by the line:

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