# Question #e6a68

##### 1 Answer

#### Explanation:

You can actually solve this problem without doing any calculations.

Start by taking a look at the **ideal gas law** equation

#color(blue)(PV = nRT)" "# , where

*universal gas constant*, usually given as

**absolute temperature** of the gas

Now, the **molar volume of a gas** represents the volume occupied by **one mole** of an ideal gas under certain conditions for *pressure* and *temperature*.

Rearrange the ideal gas law equation to isolate the volume and the number of moles of gas on one side

#PV = nRT implies V/n = (RT)/P" " " "color(red)("(*)")#

Since the universal gas constant is, well, *constant*, you can write this as

#overbrace(V/n)^(color(purple)("molar volume")) prop color(white)(a)T/P#

Notice that the molar volume is **proportional** to the ratio that exists between temperature and pressure. This tells you that the molar volume will be **maximum** when the **maximum**.

One more thing to note here - it is **of the utmost importance** that you use **absolute temperature** here, i.e. temperature expressed in *Kelvin*.

So, pick one of the options as a starting point and compare it to the others. **STP** conditions are *usually* defined as a pressure of

With this as a reference point, you will have

#color(brown)("At 127"^@"C" = "400.15 K and 1 atm")#

Here the temperature **Increases**, but the pressure remains constant. This means that the **Increases** as well **bigger** under these conditions than at STP.

#color(brown)("At 0"^@"C = 273.15 K and 2 atm")#

This time, temperature remains constant and pressure **doubles**. This means that the **halved** **halved** under these conditions when compared with STP.

#color(brown)("At 273"^@"C = 546.15 K and 2 atm"#

This time, both the temperature and the pressure **increase**, but notice that they both *double*. The temperature went from

This means that the **unchanged**

Therefore, the molar volume of the gas is **maximum** at

If you want numerical proof, plug these values into equation

#n = "1 mole"#

The maximum value for

#V = (RT)/P#

will come out to be approximately