Question

A gas initially at 273k is heated such that it volume and pressure became twice their original volume what is the new temperature of the gas

Answer 1

Answer:

1092K

Explanation:

We can use the combined gas law to answer this question:

P1V1/T1 = P2V2/T2,

where P, V and T are the Pressure, Volume, and Temperature for initial (1) and Final (2) conditions.  Temperatures must be in Kelvin.

The problem states that V2 = 2V1 and P2 = 2P1.

Let's rearrange to solve for T2, which is the question:

      T2 = T1(P2/P1)(V2/V1)

Note how the pressure and temperature values are written:  as ratios.  Enter the values:

       T2 = (273K)(P2/P1)(V2/V1)

        T2 = (273K)(2P1/P1)(2V1/V1)  [Use the expressions for V2 and P2 from above]

        T2 = (273K)(2)(2)

                       T2 = 1092K