Question

Question #e6a68

Answer 1

`127^@"C"` and `"1 atm"`

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

`P` - the pressure of the gas
`V` - the volume it occupies
`n` - the number of moles of gas
`R` - the universal gas constant, usually given as `0.0821("atm" * "L")/("mol" * "K")`
`T` - the 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 `T/P` ratio is 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 `"1 atm"` and a temperature of `0^@"C"`, or `"273.15 K"`.

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 `T/P` Increases as well `->` the molar volume of the gas is 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 `T/P` ratio will be halved `->` the molar volume of the gas is 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 `"273.15 K"` to `"546.15 K"`, and the pressure went from `"1 atm"` to `"2 atm"`.

This means that the `T/P` ratio will practically remain unchanged `->` the molar volume of the gas under these conditions is approximately equal to that at STP.

Therefore, the molar volume of the gas is maximum at `127^@"C"` and `"1 atm"`.

If you want numerical proof, plug these values into equation `color(red)("(*)")` and use

`n = "1 mole"`

The maximum value for

`V = (RT)/P`

will come out to be approximately `"32.9 L"` at `127^@"C"` and `"1 atm"`.