Question

When 2grams of gaseous substance “p” is introduced to an initially evacuated flask at 25 degree Celsius , the pressure is found to be 1 atm and 3 grams of another gaseous substance “Q” is introduced in same flask. The final pressure of mixture of gases was found to be 1.5 atm. Assuming the ideal behaviour of gases, calculate the ratio of molecular masses of “p” and “Q”.

Answer 1

$\begin{aligned} m_p &= 2g\\ m_Q &= 3g\\ P_p &= 1atm\\ P_{p+Q} &= 1.5atm\\ \ P_p + P_Q &= P_{p+Q}\\ P_Q &= 0.5atm\\ M_p : M_Q &= \ ?\\ \\ \textsf{from, number of moles(n) }&= \dfrac{\textsf{mass of substance(m)}}{\textsf{molar mass(M)}}\\ \\ n_p = &\dfrac{2}{M_p}\\ \\ n_Q = &\dfrac{3}{M_Q} \end{aligned}$

The ideal gas equation is;

$PV = nRT$

For gas p and Q, their gas laws are;

$P_pV = n_pRT ---(i)\\ P_QV = n_QRT ---(i)$, respectively.

Dividing equation (i) by equation (ii), we have;

$\begin{aligned} \dfrac{P_pV}{P_QV} &= \dfrac{n_pRT}{n_QRT}\\ \\ \dfrac{P_p}{P_Q} &= \dfrac{n_p}{n_Q}\\ \\ \dfrac{1\ atm}{0.5\ atm} &= \dfrac{\frac{3}{M_p}}{\frac{2}{M_Q}}\\ \\ 2 &= \dfrac{3M_Q}{2M_p}\\ \\ 4M_p &= 3M_Q\\ \textsf{dividing both sides by }&4M_Q, \textsf{ we have;}\\ \dfrac{M_p}{M_Q} &= \dfrac{3}{4} \end{aligned}$

$\therefore$ the ratio of the molecular masses of “p” and “Q” is $3:4$.