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

`T_2=630^(\circ)C`'

Step-by-step 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* 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.