The pressure of a gas has increased from 300 kPa to 600 kPa. The volume of the container has not changed, and the number of moles of the gas has not changed. The temperature of the gas has _____.

(V1/T1) = (V2/T2)

T must be in kelvin.

To determine how the temperature of the gas has changed, we can use the Ideal Gas Law which states:

PV = nRT

Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature of the gas

In this case, the volume (V) and the number of moles (n) are given to be constant. The pressure (P) has increased from 300 kPa to 600 kPa.

Since P and V are constant, we can rewrite the Ideal Gas Law as:

(P1/T1) = (P2/T2)

Where:
P1 = initial pressure
T1 = initial temperature
P2 = final pressure
T2 = final temperature

We can plug in the provided values:

(300 kPa / T1) = (600 kPa / T2)

To isolate T2, we can cross-multiply:

300 kPa * T2 = 600 kPa * T1

Now, let's solve for T2:

T2 = (600 kPa * T1) / 300 kPa

Simplifying further:

T2 = 2T1

From the equation, we can see that the final temperature (T2) is twice the initial temperature (T1). Therefore, if the pressure of the gas has doubled, the temperature of the gas has also doubled.