Question

The raise in the temperature of a given mass of an ideal gas at constant pressure and at temperature $27$ to double its volume is.

Answer 1

From the Ideal gas law we know that,

$\frac{P_1{V}_1}{T_1}A=\frac{P_2{V}_2}{T_2}$

Please ignore the A^ in the formula

Since pressure is constant so we have,

V₁/T₁ = V₂/T₂

=> V₂/V₁ = T₂/T₁

Given temperature T₁ = 27°C = 273 + 27 = 300k

since volume becomes double so, V₂/V₁ = 2

Hence ,

T₂/300 = 2

=> T₂ = 600k = 327°C

Hence raise in temperature = 327-27 = 300°C