Given constant pressure, how does the temperature change when a $500 \cdot c m^{3}$ $500cm^3$ volume of gas at $373 \cdot K$ $373K$ is cooled so that then volume reduces to $270 \cdot c m^{3}$ $270*cm^3$ ?

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Given constant pressure, how does the temperature change when a $500 \cdot c m^{3}$ $500cm^3$ volume of gas at $373 \cdot K$ $373K$ is cooled so that then volume reduces to $270 \cdot c m^{3}$ $270*cm^3$ ?

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Answer 1 · 2017-08-28T13:59:48

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

Explanation:

We solve for $T_{2} = \frac{V_{2}}{V_{1}} \times T_{1} = \frac{270 \cdot c m^{3}}{500 \cdot c m^{3}} \times 373 \cdot K$ $T_2=V_2/V_1xxT_1=(270*cm^3)/(500*cm^3)xx373*K$

$= 201.5 \cdot K$ $=201.5*K$

Clearly the temperature has to drop to reduce the volume so substantially.

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Given constant pressure, how does the temperature change when a $500 \cdot c m^{3}$ $500cm^3$ volume of gas at $373 \cdot K$ $373K$ is cooled so that then volume reduces to $270 \cdot c m^{3}$ $270*cm^3$ ?

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Given constant pressure, how does the temperature change when a 500⋅cm3500*cm^3 volume of gas at 373⋅K373*K is cooled so that then volume reduces to 270⋅cm3270*cm^3?

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Given constant pressure, how does the temperature change when a $500 \cdot c m^{3}$ $500cm^3$ volume of gas at $373 \cdot K$ $373K$ is cooled so that then volume reduces to $270 \cdot c m^{3}$ $270*cm^3$ ?