Question:
If the granule density of potassium bicarbonate is 2.350 g/cc and the true density is 3.560 g/cc, determine the interparticle porosity of the powder.
If the granule density of potassium bicarbonate is 2.350 g/cc and the true density is 3.560 g/cc, determine the interparticle porosity of the powder.
Show Hint
Interparticle porosity is important for understanding powder flow, compressibility, and packing behavior in pharmaceutical formulations.
Updated On: Jul 14, 2026
- 0.56
- 0.44
- 0.66
- 0.34
Show Solution
Verified By Collegedunia
The Correct Option is D
Approach Solution - 1
Interparticle porosity (\(\varepsilon\)) refers to the void spaces between particles in a powder or granule bed. It can be calculated using the relationship between true density and granule (or apparent) density as follows:
\[
\varepsilon = 1 - \frac{\rho_{\text{granule}}}{\rho_{\text{true}}}
\]
Where,
\(\rho_{\text{granule}} = 2.350 \, \text{g/cc}\) (given)
\(\rho_{\text{true}} = 3.560 \, \text{g/cc}\) (given)
Substitute values: \[ \varepsilon = 1 - \frac{2.350}{3.560} = 1 - 0.659 = 0.341 \] Therefore, the interparticle porosity is approximately 0.34.
\(\rho_{\text{true}} = 3.560 \, \text{g/cc}\) (given)
Substitute values: \[ \varepsilon = 1 - \frac{2.350}{3.560} = 1 - 0.659 = 0.341 \] Therefore, the interparticle porosity is approximately 0.34.
Was this answer helpful?
0
0
Show Solution
Verified By Collegedunia
Approach Solution -2
Interparticle porosity is the fraction of a granule's total volume that is empty void space, rather than solid material. Working this out per gram of powder, using specific volumes instead of density directly, checks whether each option matches the given data.
- 0.56: Converting both densities into specific volume (volume per gram) and taking the void fraction from those, worked out below, gives a porosity far smaller than 0.56, so this value is too high to match the given densities.
- 0.44: This is also higher than the porosity that the specific volume calculation below produces, so it does not fit the given densities either.
- 0.66: This value implies that nearly two-thirds of the granule's volume is empty space, which is far more voidage than densities this close together, 2.350 g/cc versus 3.560 g/cc, would produce.
- 0.34: Taking the specific volume of one gram of granules, \(1/2.350 = 0.4255\) cc, and the specific volume that same gram of solid would occupy with no voids at all, \(1/3.560 = 0.2809\) cc, the difference between them, \(0.4255 - 0.2809 = 0.1446\) cc, is the void volume per gram. Dividing this void volume by the total granule volume per gram, \(0.1446 / 0.4255\), gives about 0.34, meaning roughly 34 percent of the granule's volume is empty space. This matches the given densities directly.
Since the void volume per gram divided by the total granule volume per gram works out to about 0.34, this is the porosity that fits the given densities.
So the correct answer is 0.34.
Was this answer helpful?
0
0
Top GPAT Pharmaceutics Questions
- Which of the following is an example of a thermosetting plastic material?
- Which of the following vehicles (not required to be sterile, but must be pyrogen free) is intended to be used in the manufacture of injectable products to be sterilized after preparation?
A typical skin cream consisting of stearic acid, potassium hydroxide, glycerin, water, preservative and perfume, would be commonly known as:
- Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: The pressure filling method for filling pharmaceutical aerosol is usually preferred over cold filling method
Reason R: Because with pressure filling method high production speed can be achieved and less propellant is lost.
In light of the above statements, choose the correct answer from the options given below - Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: In the manufacturing of glycero‐gelatin suppositories, overfilling of mould is necessary.
Reason R: Glycero‐gelatin bases contract very little on cooling and the excess cannot be neatly removed.
In light of the above statements, choose the correct answer from the options given below
View More Questions
Top GPAT Questions
- Match the symbols in the List I with the terms used in the List II of conductance measurements, and
List I
SymbolsList II
TermsA \(\Omega^{-1}\) I Specific conductance B \(∧\) II Electrical conductance C k III Specific resistance D \(\rho\) IV Equivalent conductance
Choose the correct answer from the options given below: - Cottrell’s method is used for the measurement of:
- Match the statements in List I with their correct answers in List II, respectively, in respect to modified 1st law of thermodynamics, and
List I
List II
A Constant heat (q = 0) I Isothermal B Reversible process at constant temperature (dT = 0) II Isometric C Constant volume (dV = 0) III Adiabatic D Constant pressure (dP = 0) IV Isobar
Choose the correct answer from the options given below: - Boric acid with molecular weight 61.83 was partitioned between water and amyl alcohol at 25 °C. The amount of boric acid was determined to be 0.24 g in 250 ml of amyl alcohol and 0.32 g in 100 ml water. The partition coefficient of boric acid between water and amyl alcohol, when calculated at molar concentration for each of the solution, is:
- Drugs ‘A’, ‘B’ and ‘C’ follow zero order, first order and second order degradation kinetics, respectively, but have the same rate constant. Which of the following statement is true in this respect:
View More Questions