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(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.Thus,$\frac{PV}{T} = \frac{m}{M}R$.Initially,$P_1 = 10 	ext{ atm}$,$T_1 = 2

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$6$

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(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$.Thus,$\frac{PV}{T} = \frac{m}{M}R$.Initially,$P_1 = 10 	ext{ atm}$,$T_1 = 27 + 273 = 300 	ext{ K}$,and mass is $m$.So,$\frac{10V}{300} = \frac{m}{M}R \implies \frac{m}{M}R = \frac{10V}{300} = \frac{V}{30}$.After removing half the mass,the new mass is $m' = \frac{m}{2}$.The new temperature is $T_2 = 87 + 273 = 360 	ext{ K}$.Using the gas equation for the final state: $P_2 V = n' R T_2 = \frac{m}{2M} R T_2$.Substituting $\frac{m}{M}R = \frac{V}{30}$:$P_2 V = \frac{1}{2} \left( \frac{V}{30} ight) 	imes 360$.$P_2 = \frac{360}{60} = 6 	ext{ atm}$.

$A$ vessel is filled with an ideal gas at a pressure of $10 \text{ atm}$ and temperature $27^{\circ} C$. Half of the mass is removed from the vessel and the temperature of the remaining gas is increased to $87^{\circ} C$. Then the pressure of the gas in the vessel will be (in $\text{ atm}$)

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