Given data
*The given radius of a spherical container is r_1 = 14.5 cm
*The given temperature of the gas is T_1 = 179 K
*The given pressure is P_1 = 1.97 atm
*The given another pressure is P_2 = 3.73 atm
*The given another temperature is T_2 = 293 K
(A)
The radius (in cm) of the container is calculated by using the relation as
[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \frac{P_1(\frac{4}{3}\pi r^3_1)^{}}{T_1}=\frac{P_2(\frac{4}{3}\pi r^3_2)}{T_2} \\ \frac{P_1r^3_1}{T_1}=\frac{P_2r^3_2}{T_2^{}} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \frac{1.97\times(14.5)^3}{(179)}=\frac{3.73\times r^3_{2^{}}}{293} \\ r_2=\text{13}.8\text{ cm} \end{gathered}[/tex]Hence, the radius of the container is r_2 = 13.8 cm
