Ideal Gas Law and Pressure at 80°C

[SakuRERE]

Homework Statement

An ideal gas has a molar mass of 40 g and a density of 1.2 kg m-3 at 80°C. What is its pressure at that temperature?

Homework Equations

PV=nRT
R constant= 8.314
n= number of moles
T= tempreture in kelvin
density=Mass/ Volume

The Attempt at a Solution

i simply solved it like this:
i found first the volume:
V=molar mass/ density
V= 0.04kg/ 1.2kg/m^3
= 0.03 m^3
and then i just substituted in the ideal gas law

P=nRT/V (where n is the moles and here we have 1 mole)
P= (1)*(8.314)*(353K)/ (0.03) = 97828.06 Pa

but my professor solved it like this:
T = 273 + 80 = 353 K

PV = nRT

m = rV

so V = 0.04 / 1.2 with n = 1 (since 1 mole of gas has a mass of 40 g, or 0.04 kg)

P = RT/V = 8.3 x 353 x 1.2/0.04 = 8.79x104 Pa

I noticed that he didn't find the volume first and then substituted but he substituted directly in the ideal gas law! why it should be like this!?

 

[kuruman]

Homework Statement

An ideal gas has a molar mass of 40 g and a density of 1.2 kg m-3 at 80°C. What is its pressure at that temperature?

Homework Equations

PV=nRT
R constant= 8.314
n= number of moles
T= tempreture in kelvin
density=Mass/ Volume

The Attempt at a Solution

i simply solved it like this:
i found first the volume:
V=molar mass/ density
V= 0.04kg/ 1.2kg/m^3
= 0.03 m^3
and then i just substituted in the ideal gas law

P=nRT/V (where n is the moles and here we have 1 mole)
P= (1)*(8.314)*(353K)/ (0.03) = 97828.06 Pa

but my professor solved it like this:
T = 273 + 80 = 353 K

PV = nRT

m = rV

so V = 0.04 / 1.2 with n = 1 (since 1 mole of gas has a mass of 40 g, or 0.04 kg)

P = RT/V = 8.3 x 353 x 1.2/0.04 = 8.79x104 Pa

I noticed that he didn't find the volume first and then substituted but he substituted directly in the ideal gas law! why it should be like this!?

Because your professor did it right. He substituted values after he derived the algebraic expression for the unknown quantity in terms of the given quantities. Your answer is inconsistent with your professor's (and the correct) answer because you rounded the volume from 0.033333 m³ to 0.03 m³.

 

[mjc123]

And you rounded the volume to 1 significant figure, but quoted the answer to 7 sig figs! The professor gave his answer to 3 sig figs, consistent with the data provided.
Rule: do the calculations to high accuracy, round the final answer to the appropriate number of sig figs.