Question

Two closed vessels of equal volume contain air at 105 kPa, 300 K and are connected through a narrow tube.If one of the vessels is now maintained at 300 K and the other at 400 K, what will be the Pressure in the vessels ?

• A. 220 kPA
• B. 120 kPa
• C. 235 kPa
• D. 100 kPa

Answer 1

The correct option is B 120 kPa

Let the initial pressure volume and temperature in each vessel be
$P_0$ = 105 kPa), $V_0$ and $T_0$ (= 300 K)
Let the number of moles in each vessel be n. When the first vessel is maintained at temperature $T_0$ and the other is maintained at
T´ = 400 K, the pressures changes.
Let the common Pressure becomes P´ and the number of moles in the two vessels become $n_1$ and $n_2$. We have

$P_0{V}_0=nR{T}_0...\left(i\right)P^{\prime }{V}_0=n_{1}R{T}_0...\left(ii\right)P^{\prime }V0=n2R{T}^{\prime }...\left(iii\right)and{n}_1+n_2=2n....\left(iv\right)$
Putting n, $n_1$ and $n_2$ from (i), (ii) and (iii) in (iv),
$\frac{P^{\prime }{V}_0}{R{T}_0}+\frac{P^{\prime }{V}_0}{{RT}^{\prime }}=2\frac{P_0{V}_0}{R{T}_0}or,P^{\prime }\left(\frac{T^{\prime }+T_0}{T_0{T}^{\prime }}\right)=\frac{2P_0}{T_0}or,P^{\prime }=\frac{2P_0{T}^{\prime }}{T^{\prime }+T_0}=\frac{2\times 105kPa\times 400K}{400K+300K}=120kPa$