Two closed vessels of same volume are joined | vedclass

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Question

(A) Let $V$ be the volume of each vessel.Initially, the total number of moles $n = n_1 + n_2 = \frac{P_0 V}{RT_0} + \frac{P_0 V}{RT_0} = \frac{2P_0 V}

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Let $V$ be the volume of each vessel.Initially, the total number of moles $n = n_1 + n_2 = \frac{P_0 V}{RT_0} + \frac{P_0 V}{RT_0} = \frac{2P_0 V}{RT_0}$.Given $P_0 = 90 	ext{ kPa}$ and $T_0 = 400 	ext{ K}$, so $n = \frac{2 \cdot 90 \cdot V}{R \cdot 400} = \frac{180V}{400R}$.Finally, let the pressure in both vessels be $P'$ because they are connected.The total number of moles $n'$ remains constant, so $n' = n$.$n' = \frac{P' V}{RT_1} + \frac{P' V}{RT_2} = \frac{P' V}{R} (\frac{1}{400} + \frac{1}{500}) = \frac{P' V}{R} (\frac{5+4}{2000}) = \frac{9P' V}{2000R}$.Equating $n = n'$:$\frac{180V}{400R} = \frac{9P' V}{2000R}$.$P' = \frac{180}{400} \cdot \frac{2000}{9} = \frac{180}{9} \cdot \frac{2000}{400} = 20 \cdot 5 = 100 	ext{ kPa}$.

Column $- I$	Column $- II$
$(A)$ Root mean square speed of gas molecules	$(P)$ $\frac{1}{3} n m \bar{v}^{2}$
$(B)$ Pressure exerted by ideal gas	$(Q)$ $\sqrt{\frac{3 RT}{M}}$
$(C)$ Average kinetic energy of a molecule	$(R)$ $\frac{5}{2} RT$
$(D)$ Total internal energy of $1$ mole of a diatomic gas	$(S)$ $\frac{3}{2} k_{B} T$

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