Physics Q97 – AP-EAMCET 2026

Home Physics ThermodynamicsQuestion #497

An ideal gas of density '

ρ

' is enclosed in a vessel of volume

V

0^{\circ} C

temperature and

p_{0}

atmospheric pressure. If the pressure inside the vessel decreased by

Δ p

due to a leakage. The mass of the leaked out gas is

Clue 1 of 3: Ideal Gas Properties

The problem involves an ideal gas undergoing a change in pressure due to leakage while its temperature and volume remain constant.

🧠 Full Solution Path

Step 1: Relate initial density and pressure using the Ideal Gas Law✦ Active

For an ideal gas, the ideal gas law is $P V = n R T$ , where $n$ is the number of moles. The number of moles can also be expressed as $n = \frac{m}{M}$ , where $m$ is the mass and $M$ is the molar mass of the gas. So, the ideal gas law becomes $P V = \frac{m}{M} R T$ .

The initial density of the gas is given as $ρ$ . By definition, density is $ρ = \frac{m}{V}$ , which implies $m = ρ V$ . Substituting this into the ideal gas law for the initial state ( $p_{0}$ is the initial pressure):

p_{0} V = \frac{ρ V}{M} R T

We can simplify this to find a relationship between initial pressure, density, and other constants:

p_{0} = \frac{ρ R T}{M}

Rearranging this equation, we can express the term $\frac{M}{R T}$ in terms of initial density and pressure, which will be useful later:

\frac{M}{R T} = \frac{ρ}{p_{0}}

💡 Teacher's Secret Hint

Remember that the temperature $T$ should be in Kelvin when using the ideal gas law. Since it's constant, $0^{\circ} C$ ( $273.15 K$ ) is the temperature used in $R T$ in this context.

Step 2: Express initial and final mass of the gas○ Expand

The mass of the gas can be expressed from the ideal gas law $m = \frac{P V M}{R T}$ . Let $m_{0}$ be the initial mass of the gas in the vessel at pressure $p_{0}$ :

m_{0} = \frac{p_{0} V M}{R T}

Due to leakage, the pressure inside the vessel decreases by $Δ p$ . The new pressure is $p_{f} = p_{0} - Δ p$ . The volume $V$ and temperature $T$ remain constant. Let $m_{f}$ be the final mass of the gas in the vessel:

m_{f} = \frac{(p_{0} - Δ p) V M}{R T}

Step 3: Calculate the mass of the leaked gas○ Expand

The mass of the leaked gas, $m_{l e a k e d}$ , is the difference between the initial mass and the final mass:

m_{l e a k e d} = m_{0} - m_{f}

Substitute the expressions for $m_{0}$ and $m_{f}$ :

m_{l e a k e d} = \frac{p_{0} V M}{R T} - \frac{(p_{0} - Δ p) V M}{R T}

Factor out the common terms $\frac{V M}{R T}$ :

m_{l e a k e d} = \frac{V M}{R T} [p_{0} - (p_{0} - Δ p)]

Simplify the term in the brackets:

m_{l e a k e d} = \frac{V M}{R T} [Δ p]

💡 Teacher's Secret Hint

Ensure you correctly handle the subtraction of pressures. $p_{0} - (p_{0} - Δ p)$ simplifies to $Δ p$ .

Step 4: Substitute the density-pressure relationship into the leaked mass equation○ Expand

From Step 1, we established the relationship $\frac{M}{R T} = \frac{ρ}{p_{0}}$ . Substitute this into the expression for $m_{l e a k e d}$ from Step 3:

m_{l e a k e d} = V (\frac{ρ}{p_{0}}) Δ p

Rearranging the terms, we get the final expression for the mass of the leaked gas:

m_{l e a k e d} = \frac{V ρ Δ p}{p_{0}}

This matches option 4.

💡 Teacher's Secret Hint

Always check your final answer against the given options to ensure you haven't missed a simplification or made an algebraic error.

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Physics Question 97 – AP-EAMCET 2026

Clue 1 of 3: Ideal Gas Properties

🧠 Full Solution Path