A given mass of a gas occupies 143 cm3 at 27-C and 700 mm Hg pressure- What will be its volume at 300 K and 280 mm Hg pressure-

V = 143 cm^3 P = 700 mm Hg T = 27^∘C = 27 + 273 = 300 K P_1=280 mm Hg V_1 = ? T_1 = #160; 300 K Gas e ...

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Standard X

Chemistry

Gay Lussac's Law

A given mass ...

Question

A given mass of a gas occupies 143 $c m^{3}$ at $27^{\circ} C$ and $700 m m$ Hg pressure. What will be its volume at $300 K$ and $280 m m$ Hg pressure?

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Solution

V = 143 $c m^{3}$

P = $700 m m H g$

T = $27^{\circ} C$ = $27 + 273 = 300 K$

$P_{1} = 280 m m H g$

$V_{1}$ = ?

$T_{1} = 300 K$

Gas equation states that,

$\frac{P V}{T} = \frac{P_{1} V_{1}}{T_{1}}$

$\frac{700 \times 143}{300} = \frac{280 \times V_{1}}{300}$

$V_{1} = 357.5 c m^{3}$

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A given mass of a gas occupies 143 cm3 at 27∘C and 700mm Hg pressure. What will be its volume at 300K and 280mm Hg pressure?

V = 143 cm3P = 700 mm HgT = 27∘C = 27+273=300KP1=280 mm HgV1 = ?T1=300KGas equation states that,PVT=P1V1T1700×143300=280×V1300V1=357.5 cm3

A given mass of a gas occupies 143 $c m^{3}$ at $27^{\circ} C$ and $700 m m$ Hg pressure. What will be its volume at $300 K$ and $280 m m$ Hg pressure?

V = 143 $c m^{3}$

P = $700 m m H g$

T = $27^{\circ} C$ = $27 + 273 = 300 K$

$P_{1} = 280 m m H g$

$V_{1}$ = ?

$T_{1} = 300 K$

Gas equation states that,

$\frac{P V}{T} = \frac{P_{1} V_{1}}{T_{1}}$

$\frac{700 \times 143}{300} = \frac{280 \times V_{1}}{300}$

$V_{1} = 357.5 c m^{3}$