Consider a fixed volume and amount of an ideal gas at 10 °C.

What temperature would it have to be changed to in order for the pressure to be doubled?

A. 5°C

B. 20 °C

C. 566 °C

D. 293 °C

E. 278 °C

Consider a fixed volume and amount of an ideal gas at 10 °C.

What temperature would it have to be changed to in order for the pressure to be doubled?

A. 5°C

B. 20 °C

C. 566 °C

D. 293 °C

E. 278 °C

This question will be dealing with the ideal gas law equations, we can be summed up through:

**PV=nRT**

We are told that we have an ideal gas at a temperature of 10°C and we want to know what temperature increase would cause the doubling in pressure.

**Want:**

T_2

n,R, and V are constants

Leaves us with an equation of:

\frac{P_1}{T_1} = \frac{P_2}{T_2}

We want to find T_2 when we have double the pressure, so when P_2 = 2P, where P_1 = P

We also know T_1 = 10°C, which can be converted to K.

10°C = (273 + 10)K T_1 = 283K

Now sub all this into our equation:

\frac{P}{283K} = \frac{2P}{T_2}

P can be anything, it does not matter, what is important is the relationship between P_1 and P_2, which is that P_2 is double. In order to solve this easier, we can assign P a random value and then solve. P=1 is the simplest way.

\frac{1}{283K} = \frac{2(1)}{T_2}

\frac{T_2}{283K} = {2}

{T_2} = (2)(283K)

{T_2} = 566K

Now we just need to convert to °C.

$556K - 273 = 293°C

\fcolorbox{red}{grey!30}{Therefore the answer is D, $293K$.}