At a temperature T K,the pressure of 4.0 g | vedclass

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(B) Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass of argon.For the first state: $PV = \frac{

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$200$

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(B) Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass of argon.For the first state: $PV = \frac{4}{M}RT$ ... $(i)$For the second state,the temperature is $(T + 50) \ K$,the mass is $(4.0 - 0.8) = 3.2 \ g$,and the pressure $p$ remains constant: $PV = \frac{3.2}{M}R(T + 50)$ ... (ii)Equating $(i)$ and (ii) since $PV$ is constant:$\frac{4}{M}RT = \frac{3.2}{M}R(T + 50)$$4T = 3.2(T + 50)$$4T = 3.2T + 160$$0.8T = 160$$T = \frac{160}{0.8} = 200 \ K$.

At a temperature $T \ K$,the pressure of $4.0 \ g$ argon in a bulb is $p$. The bulb is put in a bath having a temperature higher by $50 \ K$ than the first one. $0.8 \ g$ of argon gas had to be removed to maintain the original pressure. The temperature $T$ is equal to ..... $K$.

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