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A quantity of an ideal gas is kept in a rigid container of constant volume. If the gas is originally at a temperature of 28 °C, at what temperature (in °C) will the pressure of the gas triple from its base value?

Respuesta :

Answer:

[tex]T_2=630^{\circ}C[/tex]'

Explanation:

Original temperature of the gas, [tex]T_1=28^{\circ}C=301\ K[/tex]

From the ideal gas equation,

[tex]P_1V_1=nRT_1[/tex]

Since,

[tex]P_2=3P_1[/tex]

[tex]nRT_2=3(nRT_1)[/tex]

[tex]T_2=3T_1[/tex]

[tex]T_2=3\times 301[/tex]

[tex]T_2=903\ K[/tex]

or

[tex]T_2=630^{\circ}C[/tex]

So, the new temperature of the gas is 630 degree Celsius. Hence, this is the required solution.

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