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PV=nRT

A sample of ethanol vapour, C2H5OH (Mr = 46.0), was maintained at a pressure of 100 kPa and at a temperature of 366 K.
State the ideal gas equation. (1 mark)
Use the ideal gas equation to calculate the volume, in cm3, that 1.36 g of ethanol vapour would occupy under these conditions. (The gas constant R = 8.31 JK–1mol–1) (4 marks)

Could anyone help out with the second part of the question?
State the ideal gas equations: PV=nRT

So what do you have?

Pressure (P): convert from kPa to Pa so 100,000 Pa
Temperature (T): 366 K
The Constant (R): 8.31
The moles (n): Use n= m/Mr so 1.36/46=0.029565

Rearrange to get volume.

V = nRT/P = 8.992 x 10^-4 m^3

We must convert from cubic metres to cubic centimetres,

Multiple by 1000x1000 to go to dm^3 then cm^3.

V = 899 cm^3
Reply 2
Avatar for Amethyst190
Amethyst190
OP
9
Original post by Miriam29
State the ideal gas equations: PV=nRT

So what do you have?

Pressure (P): convert from kPa to Pa so 100,000 Pa
Temperature (T): 366 K
The Constant (R): 8.31
The moles (n): Use n= m/Mr so 1.36/46=0.029565

Rearrange to get volume.

V = nRT/P = 8.992 x 10^-4 m^3

We must convert from cubic metres to cubic centimetres,

Multiple by 1000x1000 to go to dm^3 then cm^3.

V = 899 cm^3

so you wouldn't round the 0.029565 up to 0.03?
Original post by Amethyst190
so you wouldn't round the 0.029565 up to 0.03?

I think it depends on the exam board. Usually I’d keep all decimal places and then round with the final answer so that it is more accurate. This time there were so many decimals I stopped there.
Reply 4
Avatar for Amethyst190
Amethyst190
OP
9
Original post by Miriam29
I think it depends on the exam board. Usually I’d keep all decimal places and then round with the final answer so that it is more accurate. This time there were so many decimals I stopped there.

Okay thank you!

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