Question

$0.40g$ of helium in a bulb at a temperature of $TK$ had a pressure of $p$ atm. When the bulb was immersed in water bath at temperature $50K$ more than the first one, $0.08g$ of gas had to be removed to restore the original pressure. The temperature $T$ is:

• A. $500K$
• B. $400K$
• C. $600K$
• D. $200K$

Answer 1

The correct option is D $200K$

$PV=nRT=\frac{w}{M}RT$

$P=\frac{w}{VM}RT$

$wT=w^{\prime }{T}^{\prime }$
$w$ and $T$ are initial mass and initial temperature respectively.
$w^{\prime }$ and $T^{\prime }$ are final mass and final temperature respectively.
$0.40T=(0.40-0.08)\times (T+50)$
$0.40T=0.32T+16$

$0.08T=16$
$T=200K$