Question

A 21.6 ml ball has a pressure of 879.3 torr at 331 K. If the temperatire spikes 410 K and the pressure increases to 956.2 torr what is the volume of the bal

Answer 1

Answer:

24.6 mL is the volume of the ball when the temperature spikes 410 K and the pressure increases to 956.2 Torr

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure inside a ball  = 879.3 Torr

$P_2$ = final pressure inside a ball = 956.2 Torr

$V_1$ = initial volume of ball = $21.6 mL$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $331 K$

$T_2$ = final temperature of gas = $410 K$

Now put all the given values in the above equation, we get:

$\frac{879.3 Torr \times 21.6 mL}{331 K}=\frac{956.2 Torr\times V_2}{410 K}$

$V_2=24.6 mL$

24.6 mL is the volume of the ball when the temperature spikes 410 K and the pressure increases to 956.2 Torr