Question

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?

Answer 1

Answer:

$T_2=630^{\circ}C$'

Explanation:

Original temperature of the gas, $T_1=28^{\circ}C=301\ K$

From the ideal gas equation,

$P_1V_1=nRT_1$

Since,

$P_2=3P_1$

$nRT_2=3(nRT_1)$

$T_2=3T_1$

$T_2=3\times 301$

$T_2=903\ K$

or

$T_2=630^{\circ}C$

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