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NEET Physics QUESTION #1040
Question 1
A container with \(V_1=2\ \text{L},\ n_1=5\ \text{mol},\ p_1=1\ \text{atm}\) and \(V_2=3\ \text{L},\ n_2=4\ \text{mol},\ p_2=2\ \text{atm}\) is separated by a thermal insulator. When partition is removed, equilibrium pressure is:
  • 1.4 atm
  • 1.8 atm
  • 1.3 atm
  • 1.6 atm✔️
Correct Answer Explanation
Since the partition is a thermal insulator, after mixing the total internal energy is conserved. For ideal gas: \(U \propto nT\). From ideal gas law: \(T_1=\frac{p_1V_1}{n_1R}=\frac{1\times2}{5R}=\frac{2}{5R}\) and \(T_2=\frac{2\times3}{4R}=\frac{6}{4R}\). Final \(T_f = \frac{n_1T_1+n_2T_2}{n_1+n_2}=\frac{5\cdot\frac{2}{5R}+4\cdot\frac{6}{4R}}{9}=\frac{\frac{2}{R}+\frac{6}{R}}{9}=\frac{8}{9R}\). Final \(p=\frac{(n_1+n_2)RT_f}{V_1+V_2}=\frac{9R\cdot\frac{8}{9R}}{5}=\frac{8}{5}=1.6\ \text{atm}\).