What Volume of air will be expelled from a vessel containing 250cm3 of it at constant pressue 10 ordm-C- when it is heated to 40 ordm-C at the same pressure - nbsp-

V1 = 250 cm3 T1 = 10 + 273 = 283 K T2 nbsp;= 40 + 273 = 313 K According to charles law : V1T1 #160;= #160;V2T2250283 #160;= #160;V2313V2 #160;= #1 ...

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Standard XII

Physics

Thermodynamic Processes

What Volume o...

Question

What Volume of air will be expelled from a vessel containing 250cm3 of it at constant pressue 10ºC, when it is heated to 40ºC at the same pressure ?

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Solution

V₁ = 250 cm³
T₁ = 10 + 273 = 283 K
T₂= 40 + 273 = 313 K
According to charles law :
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} \frac{250}{283} = \frac{V_{2}}{313} V_{2} = \frac{250 \times 313}{283} = 276.50 c m^{3}$

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What Volume of air will be expelled from a vessel containing 250cm3 of it at constant pressue 10ºC, when it is heated to 40ºC at the same pressure ?

V1 = 250 cm3 T1 = 10 + 273 = 283 K T2 = 40 + 273 = 313 K According to charles law : V1T1 = V2T2250283 = V2313V2 = 250×313283 = 276.50 cm3

V₁ = 250 cm³
T₁ = 10 + 273 = 283 K
T₂= 40 + 273 = 313 K
According to charles law :
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} \frac{250}{283} = \frac{V_{2}}{313} V_{2} = \frac{250 \times 313}{283} = 276.50 c m^{3}$