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(C) According to the ideal gas equation,$PV = nRT$. Since $n$,$R$,and $T$ are constant,$PV = 	ext{constant}$,so $P_1V_1 = P_2V_2$. Volume of a sphere

Vedclass · Accepted Answer

$750$

Vedclass · Answer

(C) According to the ideal gas equation,$PV = nRT$. Since $n$,$R$,and $T$ are constant,$PV = 	ext{constant}$,so $P_1V_1 = P_2V_2$. Volume of a sphere is $V = \frac{4}{3} \pi r^3$. Given $r_1 = 3 \ cm$,$P_1 = 48 	imes 10^{-3} \ bar$,$r_2 = 12 \ cm$. $P_1 	imes \frac{4}{3} \pi (r_1)^3 = P_2 	imes \frac{4}{3} \pi (r_2)^3$. $P_2 = P_1 	imes (\frac{r_1}{r_2})^3 = 48 	imes 10^{-3} 	imes (\frac{3}{12})^3$. $P_2 = 48 	imes 10^{-3} 	imes (\frac{1}{4})^3 = 48 	imes 10^{-3} 	imes \frac{1}{64}$. $P_2 = \frac{48}{64} 	imes 10^{-3} = 0.75 	imes 10^{-3} \ bar$. $P_2 = 750 	imes 10^{-6} \ bar$.

$A$ spherical balloon of radius $3 \ cm$ containing helium gas has a pressure of $48 \times 10^{-3} \ bar$. At the same temperature,the pressure of a spherical balloon of radius $12 \ cm$ containing the same amount of gas will be........ $\times 10^{-6} \ bar$.

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