Question

35. A gas is heated in such a way so that is pressure and volume both becomes double. Again lowering temperature, one-fourth of initial number of moles of air has been taken in to maintain doubled volume and pressure. By what fraction, the temperature must have been raised finally? a) \( \frac{1}{5} \) times b) \( \frac{4}{5} \) times c) \( \frac{16}{5} \) times d) \( \frac{8}{5} \) times

Answer 1

Correct option is (C) \(\frac {16}5\) times

From combined gas law,

We know that,

\(\frac{P_1V_1}{T_1} = \frac{P_2 \times V_2}{T_2}\)

Here, P₂ = 2P₁ and V₂ = 2V₁

So,

\(\frac{P_1V_1}{T_1} = \frac{2P_1 \times 2V_1}{T_2}\) 

and according to the given situation

4T₁ × n₁ = T′₂n′₂ ...(at constant PV)

T₂ = 4T₁

4T₁n₁ = \(T_2 \times \frac 54 n_1 (n_2 = n_1 + \frac{n_1}4)\)

\(\therefore T'_2 = \frac{16}5 T_1\)