CBSE (Science) Class 11CBSE
Share

Books Shortlist

# Pressure of 1 G of an Ideal Gas a at 27 °C is Found to Be 2 Bar. When 2 G of Another Ideal Gas B is Introduced in the Same Flask at Same Temperature the Pressure Becomes 3 Bar. Find a Relationship Between Their Molecular Masses. - CBSE (Science) Class 11 - Chemistry

ConceptLaws of Chemical Combination Avogadro Law

#### Question

The pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at the same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

#### Solution 1

For ideal gas A, the ideal gas equation is given by,

P_AV =  n_ART  ............(i)

Where, pA and nA represent the pressure and number of moles of gas A.

For ideal gas B, the ideal gas equation is given by,

p_sV = n_sRT  ......(ii)

Where, pand nB represent the pressure and number of moles of gas B.

[V and T are constants for gases A and B]

From equation (i), we have

p_AV = m_A/M_A RT => (p_AM_A)/m_A  = (RT)/V   .....(iii)

From equation (ii), we have

p_BV = m_B/M_B RT => (p_BM_B)/m_B = RT/V .......(iv)

Where, Mand MB are the molecular masses of gases A and B respectively.

Now, from equations (iii) and (iv), we have

(p_M_A)/m_A = (p_BM_B)/m_B  .....(v)

Given

m_A = 1 g

p_A =  2 "bar"

m_B = 2g

p_B = (3 -2) = 1 "bar"

(Since total pressure is 3 bar)

Substituting these values in equation (v), we have

(2xxM_A)/1 = (1xxM_B)/2

=> 4M_A = M_B

Thus, a relationship between the molecular masses of A and B is given by

4M_A = M_B

#### Solution 2

Suppose molecular masses of A and B are MA and MB respectively. Then their number of moles will be

n_A = 1/M_A , n_B= 2/M_B

P_A =  "2bar", P_A + P_B =  3 bar, i.e P_B =  1 "bar"

Applying the relation PV = nRT

P_AV = n_ART, P_BV = n_BRT

:. P_A/P_B = n_A/n_B = (1/M_A)/(2/M_B) = M_B/(2M_A)

or M_B/M_A =  2xx P_A/P_B = 2 xx 2/1  =\ 4 or M_B =  4 M_A

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution Pressure of 1 G of an Ideal Gas a at 27 °C is Found to Be 2 Bar. When 2 G of Another Ideal Gas B is Introduced in the Same Flask at Same Temperature the Pressure Becomes 3 Bar. Find a Relationship Between Their Molecular Masses. Concept: Laws of Chemical Combination - Avogadro Law.
S