A gas at $34$ $34$ $^{\circ} C$ $""^@C$ expressed a volume of $10 \cdot L$ $10*L$ in a piston; what is the new volume at $48$ $48$ $^{\circ} C$ $""^@C$ ?

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A gas at $34$ $34$ $^{\circ} C$ $""^@C$ expressed a volume of $10 \cdot L$ $10*L$ in a piston; what is the new volume at $48$ $48$ $^{\circ} C$ $""^@C$ ?

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Answer 1 · 2017-05-13T03:53:42

Old Charles' Law holds that $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$ , moles and pressure constant.........

Explanation:

Pressure is a constant here, but we know (or should know) that $1 \cdot a t m$ $1*atm$ pressure will support a column of mercury that is $760 \cdot m m \cdot H g$ $760*mm*Hg$ tall; this relationship is linear: 1/2 an atmosphere will support a column of mercury that is $380 \cdot m m \cdot H g$ $380*mm*Hg$ tall. A mercury manometer is thus useful to measure laboratory pressures of confined gases.

We also must employ the absolute temperature scale..........

And so $V_{2} = \frac{V_{1}}{T_{1}} \times T_{2} = \frac{10 \cdot L}{307 \cdot K} \times 348 \cdot K ≅ 11 \cdot L$ $V_2=V_1/T_1xxT_2=(10*L)/(307*K)xx348*K~=11*L$

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A gas at $34$ $34$ $^{\circ} C$ $""^@C$ expressed a volume of $10 \cdot L$ $10*L$ in a piston; what is the new volume at $48$ $48$ $^{\circ} C$ $""^@C$ ?

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A gas at 3434 ∘C""^@C expressed a volume of 10⋅L10*L in a piston; what is the new volume at 4848 ∘C""^@C?

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Explanation:

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A gas at $34$ $34$ $^{\circ} C$ $""^@C$ expressed a volume of $10 \cdot L$ $10*L$ in a piston; what is the new volume at $48$ $48$ $^{\circ} C$ $""^@C$ ?