Question

An ideal gas raises in temperature from 102.0 to 212.0 ^oC and the pressure increases from 49200 to 95800 Pa. What are the initial and final densities of that gas in mol/L?

Answer 1

The ideal gas law can be used to make these calculations. PV=nrt so n/v= P/(R)(T). If we want to use R = 0.0821 L-atm/K-mol then we need to convert the two pressures to atmospheres and we must convert temperature to degrees Kelvin (C + 273);

102 C + 273 = 375K and (49200Pa)(1 a2m/101325 Pa) = 0/4856 atm.

So n/V = (0.4856 atm/(0.0821)(375) = 0.0158 mol/L.

For the second condition, T = 212 + 273 = 485K and P = 95800/101325 = 0.9455 atm.

Again, n/V = P/(RT) = 0.9455/(0.0821)(485) = .02374 mol/L.