Assuming the same pressure in each case, calculate the mass of hydrogen required to inflate a balloon to a certain volume V at $100^{o}$ C if $3.5$ g helium is required to inflate the balloon to half the volume, $0.50$ V, at $25^{o}$ C.

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Solution

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

$\frac{n_{1} \times R \times 373}{V} = \frac{3.5 \times R \times 298}{4 \times 0.5 V}$

$n_{1} = \frac{3.5 \times 298}{2 \times 373}$

$n_{1} = 1.4$

$\frac{w t}{2} = 1.4$

$W e i g h t o f H_{2} = 2.8 g m$

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Assuming the same pressure in each case, calculate the mass of hydrogen required to inflate a balloon to a certain volume V at 100oC if 3.5 g helium is required to inflate the balloon to half the volume, 0.50V, at 25oC.

n1R1T1V1=n2R2T2V2n1×R×373V=3.5×R×2984×0.5Vn1=3.5×2982×373n1=1.4wt2=1.4WeightofH2=2.8 gm

Assuming the same pressure in each case, calculate the mass of hydrogen required to inflate a balloon to a certain volume V at $100^{o}$ C if $3.5$ g helium is required to inflate the balloon to half the volume, $0.50$ V, at $25^{o}$ C.

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

$\frac{n_{1} \times R \times 373}{V} = \frac{3.5 \times R \times 298}{4 \times 0.5 V}$

$n_{1} = \frac{3.5 \times 298}{2 \times 373}$

$n_{1} = 1.4$

$\frac{w t}{2} = 1.4$

$W e i g h t o f H_{2} = 2.8 g m$