99 Percentile Qs Bank for JEE MainChemistryThermodynamics (C)
A rigid and insulated tank of 3   m 3 volume is divided into two compartments. One compartment of volume of 2   m 3 contains an ideal gas at 0 . 8314   MPa and 400   K and while the second compartment of volume 1   m 3 contains the same gas at 8 . 314   MPa and 500   K . If the partition between the two compartments is ruptured, the final temperature of the gas is
Options
- A420   K
- B450   K
- C480   K
- DNone of these
Correct answer
C. 480   K
Step-by-step solution
Moles of the gas in the first compartment: n 1 = P 1 V 1 RT 1 = 0.8314 × 1 0 6 × 2 8.314 × 4 0 0 = 5 0 0 Similarly, n 2 = 2000 The tank is a rigid and insulated, hence, w = 0 and q = 0 . Therefore, Δ U = 0 . Let T f and P f denote the final temperature and pressure, respectively. Δ U = n 1 C V, m T f - T 1 + n 2 C V, m T f - T 2 = 0 5 0 0 T f - 400 + 2 0 0 0 T f - 500 = 0 T f = 480 K