Thermodynamics — JEE Main practice

Step 1: Find the heat required per second to warm the water

The heat energy needed to raise the temperature of a mass is $Q = mc\Delta T$, so the required heating power is: $P_{req} = \frac{dQ}{dt} = \frac{dm}{dt}\cdot c\cdot\Delta T$

Given: water flows at 5.0 litres/min. Since the density of water is 1 kg/L, this is a mass flow rate of: $\frac{5.0}{60}\text{ kg/s}$

Temperature change: $\Delta T = 87-27 = 60^{\circ}\text{C}$

Specific heat: $c = 4200\text{ J/kg·}^{\circ}\text{C}$

$P_{req} = \left(\frac{5.0}{60}\right)\times4200\times60 = 21000\text{ J/s}$

Step 2: Set up the heat produced by burning gas

Let $R$ be the rate of gas consumption in g/s. The heat produced per second is: $P_{prod} = R\times H$

where $H = 5.0\times10^4\text{ J/g}$ is the heat of combustion.

$P_{prod} = R\times50000\text{ J/s}$

Step 3: Equate heat produced to heat required (assuming 100% efficiency)

$R\times50000 = 21000$

$R = \frac{21000}{50000} = 0.42\text{ g/s}$

This matches option B. This means the geyser must burn gas at a steady rate of 0.42 grams every second to keep up with heating the flowing water continuously.

Background: RMS speed formula

According to kinetic theory, the root-mean-square speed of gas molecules is: $v_{rms} = \sqrt{\frac{3RT}{M}}$

where $T$ is absolute temperature and $M$ is molar mass.

Step 1: Set up the equality condition

We're told the rms speeds of oxygen and hydrogen are equal: $v_{rms(O_2)} = v_{rms(H_2)}$

$\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$

Squaring both sides and cancelling $3R$: $\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$

Step 2: Substitute the known values

Temperature of oxygen: $T_{O_2} = 47+273 = 320\text{ K}$

Molar masses: $M_{O_2} = 32$, $M_{H_2} = 2$ (ratio given as 32/2)

$\frac{320}{32} = \frac{T_{H_2}}{2}$

Step 3: Solve for $T_{H_2}$

$10 = \frac{T_{H_2}}{2} \implies T_{H_2} = 20\text{ K}$

Step 4: Convert to Celsius

$T_{H_2}(^{\circ}\text{C}) = 20-273 = -253^{\circ}\text{C}$

This matches option C. This result makes sense: since hydrogen molecules are much lighter than oxygen molecules, they naturally move faster at the same temperature — so to slow them down to match oxygen's speed at a relatively warm 47°C, hydrogen must be cooled to a very low temperature.

Background: Adiabatic process relation

Since $\Delta Q = 0$, this is an adiabatic process. For a reversible adiabatic process, temperature and volume are related by: $T\cdot V^{\gamma-1} = \text{constant}$

where $\gamma = \dfrac{C_p}{C_v}$ depends on the gas's atomicity.

Step 1: Set up the relation between initial and final states

$T_1\cdot V_1^{\gamma-1} = T_2\cdot V_2^{\gamma-1}$

Given: $V_2 = 8V_1$ and $T_2 = \dfrac{T_1}{4}$.

$T_1\cdot V_1^{\gamma-1} = \left(\frac{T_1}{4}\right)\cdot(8V_1)^{\gamma-1}$

Step 2: Solve for $\gamma$

$V_1^{\gamma-1} = \frac{1}{4}\cdot8^{\gamma-1}\cdot V_1^{\gamma-1}$

$1 = \frac{1}{4}\cdot8^{\gamma-1} \implies 4 = 8^{\gamma-1}$

Writing both sides as powers of 2: $2^2 = 2^{3(\gamma-1)}$

$2 = 3(\gamma-1) \implies 2 = 3\gamma-3 \implies 3\gamma = 5 \implies \gamma = \frac{5}{3}$

Step 3: Relate $\gamma$ to degrees of freedom

$\gamma = 1+\frac{2}{f}$

$\frac{5}{3} = 1+\frac{2}{f} \implies \frac{2}{3} = \frac{2}{f} \implies f=3$

Step 4: Identify which gas has 3 degrees of freedom

A gas with $f=3$ has only translational motion (no rotation or vibration activated) — this is characteristic of a monoatomic gas.

Checking the options:

• $NH_3$ (ammonia): polyatomic (non-linear) — not monoatomic
• $O_2$: diatomic — not monoatomic
• $CO_2$: polyatomic (linear) — not monoatomic
• He (helium): a noble gas that exists as single atoms — monoatomic

Therefore, the gas must be He, matching option B.