Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Thermodynamics

Question:

At 10°C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110°C this ratio is : 

Options:

\(\frac{283}{383}x\)

\(\frac{383}{283}x\)

x

\(\frac{10}{110}x\)

Correct Answer:

\(\frac{283}{383}x\)

Explanation:

PV = nRT 

⇒ \(P\frac{m}{\rho} = \frac{m}{M}RT\)

⇒ \(\frac{(\frac{\rho}{P})_f}{(\frac{\rho}{P})_i} = \frac{T_i}{T_f} = \frac{10 + 273}{110 + 273} = \frac{283}{383}\)

⇒ \(\frac{(\frac{\rho}{P})_f}{X} = \frac{283}{383}\)

∴\((\frac{\rho}{P})_f = \frac{283}{383} X\)