A balloon contains 500 m^3 of helium at 27^{ | vedclass

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(C) Using the ideal gas law in the form of the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.Given values are $P_1 = 1 \ atm$,$V_1 = 5

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$900$

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(C) Using the ideal gas law in the form of the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.Given values are $P_1 = 1 \ atm$,$V_1 = 500 \ m^3$,and $T_1 = 27 + 273 = 300 \ K$.Final state values are $P_2 = 0.5 \ atm$,$T_2 = -3 + 273 = 270 \ K$,and we need to find $V_2$.Substituting the values into the equation: $\frac{1 	imes 500}{300} = \frac{0.5 	imes V_2}{270}$.Solving for $V_2$: $V_2 = \frac{500 	imes 270}{300 	imes 0.5} = \frac{500 	imes 270}{150} = 500 	imes 1.8 = 900 \ m^3$.

$A$ balloon contains $500 \ m^3$ of helium at $27^{\circ}C$ and $1 \ atm$ pressure. The volume of the helium at $-3^{\circ}C$ temperature and $0.5 \ atm$ pressure will be ..... $m^3$.

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