JEE Main 2016 · 3 Apr · Shift 1 · Chemistry Q32

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ChemistrySingle correct+4 / -1Question 32 of 90

Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure $p_i$ and temperature $T_1$ are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to $T_2$ . The final pressure $p_f$ is :

Two bulbs of volume V joined by a narrow tube, both at T₁ and pressure pᵢ; afterwards one is at T₁ and the other at T₂, both at pressure p_f.

p_i\left(\dfrac{T_1T_2}{T_1+T_2}\right)

2p_i\left(\dfrac{T_1}{T_1+T_2}\right)

2p_i\left(\dfrac{T_2}{T_1+T_2}\right)

2p_i\left(\dfrac{T_1T_2}{T_1+T_2}\right)

Answer and worked solution

Official answer: C

The total amount is fixed: $\dfrac{2p_iV}{RT_1}=\dfrac{p_fV}{RT_1}+\dfrac{p_fV}{RT_2}$ , so
$p_f=2p_i\left(\frac{T_2}{T_1+T_2}\right).$

Topics:Gas laws

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JEE Main 2016 · 3 Apr · Shift 1, Chemistry question 32