Gaseous State (JEE Advanced)

Question 1

An open flask contains air at 27 °C and 1.00 atm. It is heated at constant (atmospheric) pressure until one-fourth of the moles of air originally present have been expelled. The flask is then stoppere

Accepted Answer

C. At constant P and V, nT = constant. Remaining moles = 3/4 n₀, so T = 300 × 4/3 = 400 K = 127 °C.
After stoppering, n and V are fixed: P = 1.00 × 300/400 = 0.75 atm.
Traps: 36 °C uses °C in place of K (27 × 4/3); 400 °C forgets to convert back; 102 °C / 0.80 atm treats 'one-fourth expelled' as 'volume increased by one-fourth' (T = 375 K).
Difficulty: Hard

Question 2

For a fixed amount of an ideal gas, consider the usual gas-law plots. Which statements are correct? One or more options may be correct.

Accepted Answer

A, C. PV = nRT.
(A) True: V = (nR)/P(t + 273.15), so every isobar meets V = 0 at t = −273.15 °C, the basis of the absolute scale.
(B) False: P = (nR)/VT; slope nR/V is smaller for a larger volume.
(C) True: P = nRT·1/V; slope nRT ∝ T, line passes through the origin.
(D) False: V = (nR)/PT; slope nR/P decreases with pressure.
Difficulty: Hard

Question 3

Two glass bulbs of volumes 1.00 L and 3.00 L are joined by a tube of negligible volume and contain an ideal gas at 300 K and 1.00 atm. The 1.00 L bulb is then heated to 450 K while the 3.00 L bulb is

Accepted Answer

A. Total moles are conserved; the final pressure P is common to both bulbs.
(P₀(V₁+V₂))/(T₀) = P((V₁)/(T₁) + (V₂)/(T₂))
(1.00 × 4.00)/300 = P(1.00/450 + 3.00/250), so 0.013333 = P(0.0022222 + 0.012000) = 0.014222 P and P = 0.9375 atm (= 15/16).
Fraction in the hot bulb = (1/450)/(1/450 + 3/250) = 0.0022222/0.014222 = 0.15625 = 15.6%.
Trap: the volume-weighted mean temperature is exactly 300 K, wh

Gaseous State: questions with solutions

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