Exam practice question

Question

I need help solving this problem I am pretty sure it has to do with the Ideal Gas Law, but I don't understand how to solve cause in the Ideal gas law you don't use density or in my understanding you don't.

Q.) You've devised a helium thermometer for extreme temperatures where the density alone can be used to determine the temperature. Find the temperature, in Celsius, based on this premise if the density of helium gas is 0.0800 g/L at 1.00 atm.

a. 946 ^oC

b. 609 ^oC

c. 336 ^oC

d. 340 ^oC

Answer 1

Since PV = nRT then T = (PV)/(nR) = (P/R)×(V/n)

where P = 1.00 atm (given) and R = 0.08206 (L-atm)/(mol-K)

Other units are n = mol and V = L

We are given that the density of helium gas is 0.0800 g/L

To use T = (P/R)×(V/n) to calculate temperature we want density in mol/L

Helium has an atomic mass of 4 g/mol, so

ρ_He =.08 g/L × 1/4 mol/g = 0.02 mol/L = n/V for helium

This means that (V/n) in our equation for temperature is given by

V/n = 1/(0.02 mol/L) = 50 L/mol

T (in °K) is given by

T = (P/R)×(V/n)

where (P/R) = (1/0.08206) and (V/n) = 50

So T = 50/0.08206 = 609.3 °K

T (in °C) = T (in °K) – 273 = 609.3 – 273 = 336.3

Answer

c. 336 °C