A balloon contains 500 m 3 of helium at 27- C and 1 atmosphere pressure- The volume of the helium at -3- C temperature and 0-5 atmosphere pressure will beA- 700 m 3B- 500 m3C- 900 m 3D- 1000 m 3

The correct option is C From PV=μ RT we get V_2/V_1= ( T_2/T_1 ) ( P_1/P_2 )= ( 270/300 ) ( 1/0.5 )=9/5⇒ V_2=500×9/5=900 m^3 ...

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Standard XII

Chemistry

Pressure and Temperature

A balloon con...

Question

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

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Solution

The correct option is C
From $P V = μ R T$ we get $\frac{V_{2}}{V_{1}} = (\frac{T_{2}}{T_{1}}) (\frac{P_{1}}{P_{2}}) = (\frac{270}{300}) (\frac{1}{0.5}) = \frac{9}{5} \Rightarrow V_{2} = 500 \times \frac{9}{5} = 900 m^{3}$

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A balloon contains 500m3 of helium at 27∘C and 1 atmosphere pressure. The volume of the helium at –3∘C temperature and 0.5 atmosphere pressure will be

The correct option is C From PV=μRT we get V2V1=(T2T1)(P1P2)=(270300)(10.5)=95⇒V2=500×95=900m3

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The correct option is C
From $P V = μ R T$ we get $\frac{V_{2}}{V_{1}} = (\frac{T_{2}}{T_{1}}) (\frac{P_{1}}{P_{2}}) = (\frac{270}{300}) (\frac{1}{0.5}) = \frac{9}{5} \Rightarrow V_{2} = 500 \times \frac{9}{5} = 900 m^{3}$