Question

A container X has volume double that of a container Y and both are connected by a thin tube. Both contain same ideal gas. The temperature of X is 200K and that of Y is 400K. If the mass of gas in X in m then in Y it will be:

• A. m/8
• B. m/6
• C. m/4
• D. m/2

Answer 1

The correct option is C m/4

Ideal gas equation $PV=nRT$.
For gas X: $P=P,V=2V,(n=m/a);T=200K$.
Final position: $P=P,V=V,(n=m_2/a),T=400K$.
By comparing them, $m_2=\left(m/4\right)$.