Two closed containers of equal volume filled with air at pressure P0 and temperature T0. Both are connected by a narrow tube. If one of the container is maintained at temperature T0 and other at temperature T, then new pressure in the containers will be

Question

Two closed containers of equal volume filled with air at pressure P0 and temperature T0. Both are connected by a narrow tube. If one of the container is maintained at temperature T0 and other at temperature T, then new pressure in the containers will be (A) (2 P0 T/T+T0) (B) (P0 T/T+T0) (C) (P0 T/2(T+T0)) (D) (T+T0/P0)

Tardigrade Admin · Accepted Answer

(P/T)= constant Initially (P0/T0)+(P0/T0)=(2 P0/T0) For two containers (P0/T0)+(P/T)=(2 P0/T0) P=(2 P0 × T × T0/T0(T+T0)) or P=(2 P0 T/(T+T0))

Q. Two closed containers of equal volume filled with air at pressure $P_{0}$ and temperature $T_{0}$ . Both are connected by a narrow tube. If one of the container is maintained at temperature $T_{0}$ and other at temperature $T$ , then new pressure in the containers will be

$\frac{2 P _{0} T}{T + T _{0}}$

$\frac{P _{0} T}{T + T _{0}}$

$\frac{P _{0} T}{2 ( T + T _{0} )}$

$\frac{T + T _{0}}{P _{0}}$

Q. Two closed containers of equal volume filled with air at pressure P0​ and temperature T0​. Both are connected by a narrow tube. If one of the container is maintained at temperature T0​ and other at temperature T, then new pressure in the containers will be

T+T0​2P0​T​

T+T0​P0​T​

2(T+T0​)P0​T​

P0​T+T0​​

TP​= constant Initially T0​P0​​+T0​P0​​=T0​2P0​​ For two containers T0​P0​​+TP​=T0​2P0​​ P=T0​(T+T0​)2P0​×T×T0​​ or P=(T+T0​)2P0​T​

Q. Two closed containers of equal volume filled with air at pressure $P_{0}$ and temperature $T_{0}$ . Both are connected by a narrow tube. If one of the container is maintained at temperature $T_{0}$ and other at temperature $T$ , then new pressure in the containers will be

$\frac{2 P _{0} T}{T + T _{0}}$

$\frac{P _{0} T}{T + T _{0}}$

$\frac{P _{0} T}{2 ( T + T _{0} )}$

$\frac{T + T _{0}}{P _{0}}$