Question

A certain gas has a density of 1.30 g/L at STP. What will its density, in g/L, be at 546 K and 380 torr pressure ?

Answer 1

The gas will have the same density at 546 K and 380 torr that it had at STP.

There are two ways to prove this - a longer one and a shorter one. Here's the longer one.

Standard Temperature and Pressure - STP - is defined as having a temperature of `"273.15 K"` and a pressure of `"1.00 atm"`. You can use the ideal gas law equation for the two states at which the gas is found - for STP (1) and for the second state (2) (not-STP). So,

`P_1V_1 = nRT_1` - the number of moles of gas does not change;

The number of moles is defined as the ratio between the gas' mass and its molar mass; plug this equation into the above one

`n = m/M_m => P_1V_1 = m/M_m * RT_1 = m * T_1 * R/M_m`

Solve for `R/M_m`, which is a constant (product of two constants)

`R/M_m = (P_1V_1)/(m * T_1)`

Density is defined as mass per unit of volume, `rho = m/V`; if you look closely, you'll notice that the above equation has `V_1/m`, which is equal to `1/(rho)`. Plug this into the equation to get

`R/M_m = P_1/T_1 * 1/(rho_1)`

If you write the equation for the non-STP state, you'll get the exact same thing ,except you'll now have `rho_2`

`R/M_m = P_2/T_2 * 1/(rho_2)`

This means that

`P_1/T_1 * 1/(rho_1) = P_2/T_2 * 1/(rho_2)`

In you case,

`"1.00 atm"/"273 K" * 1/"1.30 g/L" = ((760/380)"atm")/"546 K" * 1/(rho_2) => rho_2 = "1.30 g/L"`

Now for the shorter answer. Notice that the pressure drops by half and the temperature doubles; if you take into account the fact that the number of moles is constant, the obvious conclusion is that volume remains constant as well.

SInce density is mass per volume, and neither mass nor volume change, then the density will be the same for both the STP and the non-STP conditions.