# Question #5af4f

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that because the pressure and the number of moles of gas are being **kept constant**, the volume of the gas will have a **direct relationship** to its temperature, as described by **Charles' Law**.

Simply put, when the temperature of the **increases**, the volume of the gas **increases** by the same factor. Similarly, when the temperature of the gas **decreases**, the volume **decreases** by the same factor.

Mathematically, this can be written as

#color(blue)(ul(color(black)(V_1/T_1 = V_2/T_2)))#

Here

#V_1# and#T_1# represent the volume and the temperature of the gas at an initial state#V_2# and#T_2# represent the volume and the temperature of the gas at a final state

Before moving on, make sure that you **convert** the temperature from *degrees Celsius* to *Kelvin*

#T = 60^@"C" + 273.15 = "333.15 K"#

Now, you know that the volume of the gas must **triple** as a result of the increase in temperature. Even without setting up the equation, you should be able to say that in order for that to happen, the temperature **must triple** as well.

So

#T_2 = 3 * T#

#T_2 = 3 * "333.15 K" = "999.45 K"#

You can show that this is the case by using the equation. You know that

#V_2 = 3 * V_1#

Rearrange the equation to solve for

#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#

Plug in your values to find

#T_2 = (3 * color(red)(cancel(color(black)(V_1))))/color(red)(cancel(color(black)(V_1))) * "333.15 K" = "999.45 K"#

Rounded to one **significant figure**, the number of sig figs you have for the initial temperature of the gas, the answer will be

#color(darkgreen)(ul(color(black)(T_2 = "1000 K")))#

If you want, you can express this in *degrees Celsius*

#t_2[""^@"C"] = "999.45 K" - 273.15 = 726.3^@"C"#

Ince again, you must round the answer to one significant figure

#color(darkgreen)(ul(color(black)(t_2 = 700^@"C")))#