Solveeit Logo
Login

Question

Question: 7.5 l of the particular gas at stp based 16 g what is the vapour density of gas pressure is equal to...

7.5 l of the particular gas at stp based 16 g what is the vapour density of gas pressure is equal to 1 bar

Answer

24.21

Explanation

Solution

To determine the vapor density of the gas, we first need to find its molar mass. The problem specifies that the conditions are STP, and the pressure is 1 bar.

  1. Define STP (Standard Temperature and Pressure) with 1 bar pressure: According to IUPAC, STP is defined as 0 °C (273.15 K) and 1 bar (100 kPa) pressure.

  2. Determine the molar volume of an ideal gas at this STP: Using the ideal gas law, PV=nRTPV = nRT, the molar volume Vm=RTPV_m = \frac{RT}{P}. Given R=0.08314 L bar K−1 mol−1R = 0.08314 \text{ L bar K}^{-1} \text{ mol}^{-1}, T=273.15 KT = 273.15 \text{ K}, and P=1 barP = 1 \text{ bar}: Vm=0.08314 L bar K−1 mol−1×273.15 K1 bar≈22.7 L/molV_m = \frac{0.08314 \text{ L bar K}^{-1} \text{ mol}^{-1} \times 273.15 \text{ K}}{1 \text{ bar}} \approx 22.7 \text{ L/mol}.

  3. Calculate the number of moles of the gas: We are given that 16 g of the gas occupies 7.5 L at STP. Number of moles (nn) = VolumeMolar Volume at STP\frac{\text{Volume}}{\text{Molar Volume at STP}} n=7.5 L22.7 L/mol≈0.3304 moln = \frac{7.5 \text{ L}}{22.7 \text{ L/mol}} \approx 0.3304 \text{ mol}.

  4. Calculate the molar mass of the gas: Molar mass (MM) = Mass of gasNumber of moles\frac{\text{Mass of gas}}{\text{Number of moles}} M=16 g0.3304 mol≈48.42 g/molM = \frac{16 \text{ g}}{0.3304 \text{ mol}} \approx 48.42 \text{ g/mol}.

  5. Calculate the vapor density of the gas: Vapor density (VD) is defined as half of the molar mass of the gas (assuming hydrogen gas, H2_2, with molar mass 2 g/mol, as the reference). VD=Molar mass of gasMolar mass of H2=M2VD = \frac{\text{Molar mass of gas}}{\text{Molar mass of H}_2} = \frac{M}{2} VD=48.42 g/mol2 g/mol≈24.21VD = \frac{48.42 \text{ g/mol}}{2 \text{ g/mol}} \approx 24.21.

    Note: If the gas were Ozone (O3_3), its molar mass would be 48 g/mol, and its vapor density would be 24. For 16 g of O3_3 (1/3 mol), the volume at 1 bar STP would be (1/3)×22.7 L=7.567 L (1/3) \times 22.7 \text{ L} = 7.567 \text{ L}, which is very close to the given 7.5 L. This suggests that 24 might be the intended exact answer if the input values were slightly rounded. However, based on the given numbers and standard definitions, the calculated value is 24.21.