Question

The V of a gas at 29 C and 0.53 atm is 39 mL. What V will the same gas sample occupy at standard conditions?

Answer 1

STP most commonly is used when performing calculations on gases, such as gas density. The standard temperature is $273\,\mathrm{K}$ ($0^{\circ}$ Celsius) and the standard pressure is 1 atm or $101.3\,\mathrm{kPa}$ pressure. At STP, one mole of gas occupies $22.4\,\mathrm{L}$ of volume (molar volume).

This task can be solved by using ideal gas law.

The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas law is often introduced in its common form:

$$PV = nRT$$

where $P$ is the pressure of the gas, $V$ is the volume of the gas, $n$ is the amount of substance of gas (also known as number of moles), $T$ is the temperature of the gas and $R$ is the ideal, or universal, gas constant.

So it is one amount of gas under different conditions.

$\mathrm{P_1V_1 = nRT_1}$ and,

$\mathrm{P_2V_2 = nRT_2}$

If R is constant and n is too:

$\mathrm{P_1V_1 / T_1 = nR}$ and,

$\mathrm{P_2V_2 / T_2 = nR}$, so

$\mathrm{P_1V_1 / T_1 = P_2V_2 / T_2}$

Given:

$\mathrm{T_1 = 29\,C = 302\,K}$

$\mathrm{T_2 = 273\,K}$

$\mathrm{P_1 = 0.53\,atm}$

$\mathrm{P_2 = 1\,atm}$

$\mathrm{V_1 = 39\,ml = 0.039\,L}$

$$\mathrm{P_1V_1T_2 = P_2V_2T_1}$$ $$\mathrm{V_2 = P_1V_1T_2 / T_1P_2}$$ $$\mathrm{V_2 = 0.53*0.039*302/273 = 0,023\ L\ or\ 23\ mL}$$