Question

Harvey kept a balloon with a volume of 348 milliliters at 25.0˚C inside a freezer for a night. When he took it out, its new volume was 322 milliliters, but its pressure was the same. If the final temperature of the balloon is the same as the freezer’s, what is the temperature of the freezer? The temperature of the freezer is kelvins.

Answer 1

Initial volume of the balloon = $V_{1}$= 348 mL

Initial temperature of the balloon $T_{1}$ = $25.0^{0}C + 273 = 298 K$

Final volume of the balloon $V_{2}$ = 322 mL

Final temperature of the balloon = $T_{2} = ?$

According to Charles law, volume of an ideal gas is directly proportional to the temperature at constant pressure.

$\frac{V_{1} }{T_{1} } =\frac{V_{2} }{T_{2} }$

On plugging in the values,

$\frac{348mL}{298 K} =\frac{322 mL}{T_{2} }$

$T_{2} =276 K$

Therefore, the temperature of the freezer is 276 K