Question

A perfect gas at 340 K is heated at constant pressure until its volume has increased by 18 per cent. What is the anal
temperature of the gas?​

Answer 1

Answer:

$\sf 401 \ K$

Explanation:

We have :

V₁ = V

V₂ = 18 / 100 V + V

= > V₂ = 118 / 100 V

T₁ = 340 K

According to Charle's law :

$\sf \dfrac{V_1}{T_1} =\dfrac{V_2}{T_2}$

Putting values here we get :

$\sf \dfrac{V}{340} =\left( \dfrac{\dfrac{118}{100}V }{T_2} \right)$

$\sf \dfrac{1}{340} =\left( \dfrac{\dfrac{118}{100}}{T_2} \right)$

$\sf T_2=\dfrac{118\times340}{100} \ K$

$\sf T_2=\dfrac{118\times34}{10} \ K$

$\sf T_2=401 \ K$

Hence we get required answer.