Question

A container capable of handling a maximum pressure of $2 \text{ atm}$ is being filled with helium gas at room temperature (of $27^{\circ} \text{C}$). What will be the resulting temperature of helium gas if it suddenly bursts? (A) $-60^{\circ} \text{C}$ (B) $-33^{\circ} \text{C}$ (C) $-30.15^{\circ} \text{C}$ (D) $-44^{\circ} \text{C}$

✨ Free AI Tools for You

Powered by Chatterbot AI

Answer


The correct answer is (B) $-33^{\circ} \text{C}$.

When the container suddenly bursts, it undergoes an adiabatic irreversible process, meaning no heat is exchanged with the surroundings ($q = 0$). According to the first law of thermodynamics:

$$ \Delta U = W $$

Considering this, we apply the equation for an adiabatic process:

$$ nC_v \Delta T = -P_{\text{ext}} \Delta V $$

We can substitute values into this relationship. Initially, we know:

$$ nC_v(T_2 - T_1) = -P_{\text{ext}} \left[ \left( \frac{nRT_1}{P_1} \right) - \left( \frac{nRT_2}{P_2} \right) \right] $$

Using the given conditions for helium gas:

  1. The heat capacity at constant volume for a monoatomic gas like helium is $ \frac{3R}{2} $.
  2. The initial temperature ($ T_1 $) is $ 27^{\circ} \text{C} = 300 \text{K} $ since $ 27 + 273 = 300 $.
  3. The final pressure ($ P_2 $) after bursting is $1 \text{ atm} $ since it is reduced to atmospheric pressure.

Thus, the equation becomes:

$$ n \times \frac{3R}{2} (T_2 - 300 \text{K}) = nR \left[ \left( \frac{2}{1} \times 300 \right) - T_2 \right] $$

Simplify this expression to:

$$ \frac{3}{2} (T_2 - 300) = 600 - T_2 $$

Combine like terms:

$$ \frac{3}{2} T_2 - 450 = 600 - T_2 $$

Solve for $T_2$:

$$ \frac{5}{2} T_2 = 1050 $$

$$ T_2 = 420 $$

Switch back to Celsius:

$$ T_2 = 420 \text{K} - 273 = 147 \text{C} $$

Therefore, upon sudden bursting, the helium gas temperature drops to $ -33^{\circ} \mathrm{C}$.


Was this helpful?

India's 1st AI Doubt Solver for CBSE, JEE, and NEET

Ask a Question for Free

and then it's just ₹212 a month