There is a sample of diatomic gas in a container (whose volume is constant). The temperature of this gas was increased greatly so some molecules fell into atoms (dissociated). The pressure of the gas increased by a factor of 6 and the internal energy of the gas increased to a value of 4 . 4 times the original internal energy. By what factor did the temperature of the gas (measured in Kelvin) increase?

Step-by-step solution

P V = n R T ; 6 P V = ( n 2 + 2 n 2 ) R T ′ ⇒ 6 = n 2 + 2 n 2 n T ′ T ... (i) U = n ( 5 2 R T ) ; 4.4 U = n 2 ( 5 2 R T ′ ) + 2 n 2 ( 3 2 R T ′ ) 4.4 = n ( 5 2 R T ′ ) + n 2 3 R T ′ n ( 5 2 R T ) 4.4 = n 1 + 6 5 n 2 n T ′ T ... (ii) 6 4.4 = n 1 + 2 n 2 n 1 + 6 5 n 2 = 5 n 1 + 10 n 2 5 n 1 + 6 n 2 ; 30 n 1 + 36 n 2 = 22 n 1 + 44 n 2 ; 8 n 1 = 8 n 2 ⇒ n 1 = n 2 Also, n 1 + n 2 = n ⇒ n 1 = n 2 = n 2 ; 6 = ( 1 2 + 1 ) T ′ T ⇒ T ′ = 4 T