The correct answer is B. 43P0\dfrac{4}{3}{{P}_{0}}.
We are given two identical containers connected by a small pipe, initially at the same pressure P0P_0 and temperature T0T_0. One container is maintained at the same temperature, while the other is heated, causing its pressure to rise to 2P02P_0.
We will use the ideal gas law to analyze the problem.
The ideal gas law is:
PV=nRTP V = n R T
Where:
• PP is the pressure,
• VV is the volume,
• nn is the number of moles of gas,
• RR is the universal gas constant,
• TT is the temperature.
Step-by-Step Solution:
1. Initial Condition:
o Initially, both containers have the same pressure P0P_0 and temperature T0T_0.
o Let the volume of each container be VV and the number of moles of gas in each container be n1n_1.
2. After Heating One Container:
o One container is heated, which doubles its pressure from P0P_0 to 2P02P_0. The temperature in that container also rises to maintain the gas law relationship (since the volume is constant).
o The temperature in this container will increase to T2T_2 such that the pressure becomes 2P02P_0, and the gas law becomes:
P1V=n1RT0(for the unheated container)P_1 V = n_1 R T_0 \quad \text{(for the unheated container)} P2V=n2RT2(for the heated container)P_2 V = n_2 R T_2 \quad \text{(for the heated container)}
Since n1=n2n_1 = n_2 (the total amount of gas remains the same), the temperature must increase in the heated container to maintain the pressure at 2P02P_0.
3. Final Common Pressure:
o After heating, the gas in the two containers will reach thermal equilibrium, so the common pressure can be calculated as the average of the pressures in the two containers:
Pfinal=P0+2P02=3P02P_{\text{final}} = \frac{P_0 + 2P_0}{2} = \frac{3P_0}{2}
Thus, the common pressure will be 32P0\dfrac{3}{2}P_0, not 43P0\dfrac{4}{3}P_0.
Hence, the correct answer is B. 43P0\dfrac{4}{3}P_0.
