Physics Q100 – AP-EAMCET 2026

Home Physics Kinetic TheoryQuestion #203

A vessel contains a mixture of nitrogen of mass

7 g

and carbondioxide of mass

11 g

at a temperature

290 K

and pressure

1 atm

. The density of the mixture is

Clue 1 of 3: Gas Mixture Properties

For an ideal gas mixture, the total pressure, volume, and temperature are related to the total number of moles of gas.

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Step 1: Calculate Moles of Each Gas✦ Active

First, calculate the number of moles for each gas component using their given masses and molar masses. The molar mass of Nitrogen ( $N_{2}$ ) is $28 g/mol$ , and the molar mass of Carbon Dioxide ( ${CO}_{2}$ ) is $44 g/mol$ .

n_{N_{2}} = \frac{m_{N_{2}}}{M_{N_{2}}} = \frac{7 g}{28 g/mol} = 0.25 mol

n_{{CO}_{2}} = \frac{m_{{CO}_{2}}}{M_{{CO}_{2}}} = \frac{11 g}{44 g/mol} = 0.25 mol

💡 Teacher's Secret Hint

Remember that Nitrogen exists as a diatomic molecule (\text{N}_2) when calculating its molar mass.

Step 2: Determine Total Moles and Total Mass○ Expand

Next, find the total number of moles ( $n_{t o t a l}$ ) and the total mass ( $m_{t o t a l}$ ) of the gas mixture by summing the individual components.

n_{t o t a l} = n_{N_{2}} + n_{{CO}_{2}} = 0.25 mol + 0.25 mol = 0.50 mol

m_{t o t a l} = m_{N_{2}} + m_{{CO}_{2}} = 7 g + 11 g = 18 g

Step 3: Calculate the Volume of the Gas Mixture○ Expand

Use the Ideal Gas Law, $P V = n R T$ , to calculate the total volume ( $V$ ) of the gas mixture. Given pressure ( $P = 1 atm$ ), temperature ( $T = 290 K$ ), and the calculated total moles ( $n_{t o t a l} = 0.50 mol$ ). The ideal gas constant $R = 0.0821 {L atm mol}^{- 1} K^{- 1}$ .

V = \frac{n_{t o t a l} R T}{P} = \frac{(0.50 mol) (0.0821 {L atm mol}^{- 1} K^{- 1}) (290 K)}{1 atm}

V = 11.8945 L

💡 Teacher's Secret Hint

Ensure that the units for pressure, volume, and temperature are consistent with the units of the gas constant $R$ you are using. In this case, atm, L, and K are appropriate.

Step 4: Calculate the Density of the Mixture○ Expand

Finally, calculate the density ( $ρ$ ) of the mixture by dividing the total mass ( $m_{t o t a l}$ ) by the total volume ( $V$ ). The density is $ρ = \frac{m}{V}$ .

ρ_{m i x} = \frac{m_{t o t a l}}{V} = \frac{18 g}{11.8945 L} \approx 1.5133 g/L

Rounding to three decimal places, the density is approximately $1.513 g/L$ , which is closest to option 3.

💡 Teacher's Secret Hint

Pay attention to significant figures and rounding rules, especially when comparing your calculated value to the given options.

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Physics Question 100 – AP-EAMCET 2026

Clue 1 of 3: Gas Mixture Properties

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