Questions & explanations
1. At 100 atm and 300 K, CO₂ has Z = 0.7. Its density (g/L) is closest to? (R = 0.0821 L·atm/(K·mol), M = 44 g/mol)
- 357 g/L
- 179 g/L
- 125 g/L
- 255 g/L
Answer: 179 g/L
Using real gas equation PV = ZnRT, density d = PM/(ZRT). Ideal density = PM/RT = (100×44)/(0.0821×300) ≈ 178.6 g/L. Real density = ideal/Z = 178.6/0.7 ≈ 255.1 g/L. However, note that Z = PV/(nRT) = 0.7 means the gas is more compressible; actual density = PM/(ZRT) = (100×44)/(0.7×0.0821×300) ≈ 179 g/L. The correct formula is d = PM/(ZRT).
2. Which of the following gases can be liquefied at room temperature (298 K) by applying pressure?
- Oxygen (T_c = 155 K)
- Nitrogen (T_c = 126 K)
- Hydrogen (T_c = 33 K)
- Ammonia (T_c = 405 K)
Answer: Ammonia (T_c = 405 K)
A gas can be liquefied at a given temperature if that temperature is below its critical temperature (T_c). At 298 K, only ammonia (T_c = 405 K) has T_c > 298 K, so it can be liquefied by pressure alone. Oxygen, nitrogen, and hydrogen have T_c below 298 K and cannot be liquefied at room temperature.
3. The van der Waals constants for two gases are: gas X: a = 4.17 atm L^2 mol^{-2}, b = 0.0371 L mol^{-1}; gas Y: a = 1.39 atm L^2 mol^{-2}, b = 0.0391 L mol^{-1}. Which gas is easier to liquefy and why?
- Gas X, because it has a larger a value indicating stronger intermolecular attraction.
- Gas Y, because it has a smaller a value indicating weaker intermolecular attraction.
- Gas X, because it has a smaller b value indicating smaller molecular size.
- Gas Y, because it has a larger b value indicating larger molecular size.
Answer: Gas X, because it has a larger a value indicating stronger intermolecular attraction.
Ease of liquefaction depends on the strength of intermolecular attractions, measured by the van der Waals constant a. A larger a means stronger attraction, making the gas easier to liquefy. Gas X has a = 4.17, which is larger than gas Y's a = 1.39, so gas X is easier to liquefy.
4. At constant volume and fixed amount, how does pressure of an ideal gas change with absolute temperature?
- Pressure is directly proportional to square of absolute temperature.
- Pressure is inversely proportional to absolute temperature.
- Pressure is directly proportional to absolute temperature.
- Pressure is independent of absolute temperature.
Answer: Pressure is directly proportional to absolute temperature.
Gay-Lussac's law states that at constant volume and fixed amount, pressure is directly proportional to absolute temperature (P ∝ T). This is because higher temperature increases molecular kinetic energy, leading to more forceful collisions with the container walls.
5. In the van der Waals equation (P + a n²/V²)(V - n b) = nRT, what does the term 'a n²/V²' correct for?
- The finite volume of gas molecules
- The volume occupied by the gas molecules themselves
- The increase in pressure due to molecular collisions
- The decrease in pressure due to intermolecular attractions
Answer: The decrease in pressure due to intermolecular attractions
The term a n²/V² is added to the measured pressure P to account for the fact that intermolecular attractions reduce the force with which molecules hit the walls, so the actual pressure is lower than ideal. Adding this term corrects the pressure upward.
6. N2 gas is collected over water at 27°C and 760 mm Hg. The volume is 246 mL. Aqueous tension at 27°C is 26.7 mm Hg. What is the mass of dry N2 collected? (R = 0.0821 L atm K⁻¹ mol⁻¹, N=14)
- 0.32 g
- 0.30 g
- 0.26 g
- 0.28 g
Answer: 0.28 g
Dry N2 pressure = 760 - 26.7 = 733.3 mm Hg = 733.3/760 = 0.965 atm. Use combined gas law: V_STP = (P_dry × V × T_STP)/(P_STP × T) = (0.965 × 246 × 273)/(1 × 300) = 216 mL. Moles at STP = 0.216/22.4 = 0.00964 mol. Mass = 0.00964 × 28 = 0.27 g ≈ 0.28 g.
7. At room temperature, which of the following correctly describes the Z vs P plot for H₂ and He?
- Both show Z < 1 at moderate pressures and Z > 1 at high pressures.
- H₂ shows Z < 1 at moderate pressures, while He shows Z > 1 at all pressures.
- Both show Z > 1 at all pressures.
- Both show Z < 1 at all pressures.
Answer: Both show Z > 1 at all pressures.
For H₂ and He, intermolecular attractions are negligible (small van der Waals 'a'), so the repulsive effect of finite molecular volume ('b') dominates. This causes Z = PV/nRT > 1 at all pressures at room temperature, as per NCERT Class 11 Chemistry.
8. Which value of the universal gas constant R is most appropriate when pressure is given in atm and volume in litres?
- 8.314 J K⁻¹ mol⁻¹
- 0.0821 L atm K⁻¹ mol⁻¹
- 1.987 cal K⁻¹ mol⁻¹
- 0.0831 L bar K⁻¹ mol⁻¹
Answer: 0.0821 L atm K⁻¹ mol⁻¹
According to NCERT Class 11 Chemistry, when pressure is in atm and volume in litres, the value of R is 0.0821 L atm K⁻¹ mol⁻¹. This is derived from the ideal gas equation at STP: (1 atm × 22.4 L) / (1 mol × 273.15 K) = 0.0821.
9. 250 mL of H2 is collected over water at 27°C and total pressure 750 mm Hg. Aqueous tension at 27°C is 26.7 mm Hg. What is the volume of dry H2 at STP?
- 228 mL
- 217 mL
- 206 mL
- 240 mL
Answer: 217 mL
First, find dry gas pressure: P_dry = 750 - 26.7 = 723.3 mm Hg. Then apply combined gas law: P_dry × V_collected / T_collected = P_STP × V_STP / T_STP. So V_STP = (723.3 × 250 × 273) / (760 × 300) = 216.6 mL ≈ 217 mL.
10. A gas mixture contains N2, O2 and CO2 in mole ratio 4:1:1 at 6 atm and 300 K. What is the density of the mixture in g/L? (R = 0.0821 L atm K⁻¹ mol⁻¹)
- 9.15
- 7.63
- 6.10
- 8.40
Answer: 7.63
Using ideal gas law d = PM_avg/RT. Mole fractions: N2=4/6=2/3, O2=1/6, CO2=1/6. M_avg = (2/3×28)+(1/6×32)+(1/6×44)=56/3+32/6+44/6=112/6+32/6+44/6=188/6=31.33 g/mol. d = (6×31.33)/(0.0821×300)=188/24.63=7.63 g/L.
11. What is the volume occupied by 1 mole of an ideal gas at STP as per NCERT definition?
- 24.5 L
- 22.4 L
- 22.7 L
- 11.2 L
Answer: 22.7 L
NCERT defines STP as 0°C (273.15 K) and 1 bar pressure. Using ideal gas law PV = nRT, with R = 0.08314 L bar K⁻¹ mol⁻¹, V = (1 mol × 0.08314 × 273.15) / 1 bar = 22.7 L. This is the molar volume at NCERT STP.
12. At constant temperature and pressure, what is the relationship between volume (V) and number of moles (n) of a gas?
- V ∝ √n
- V ∝ 1/n
- V ∝ n²
- V ∝ n
Answer: V ∝ n
Avogadro's law states that at constant temperature and pressure, volume is directly proportional to the number of moles (V ∝ n). Equal volumes of gases under same T and P contain equal number of molecules.