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:

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: (A) x (B) (383/283) x  (C) (10/110) x  (D) (283/383) x

Tardigrade Admin · Accepted Answer

Let the mass of the gas be m.
At a fixed temperature and pressure, volume is fixed.
Density of the gas,

ρ = \frac{m}{V}

Now

\frac{ρ}{P} = \frac{m}{P V} = \frac{m}{n RT}

\Rightarrow \frac{m}{n RT} = x

(By question)

\Rightarrow x T = constant \Rightarrow x_{1} T_{1} = x_{2} T_{2}

\Rightarrow x_{2} \Rightarrow \frac{x _{1} T _{1}}{T _{2}} = \frac{283}{383} x ⎣ ⎡ ∴ T_{1} T_{2} = = 283 K 383 K ⎦ ⎤

Q. At $1 0^{\circ}$ C the value of the density of a fixed mass of an ideal gas divided by its pressure is $x$ . At $11 0^{\circ}$ C this ratio is:

$x$

$\frac{383}{283} x$

$\frac{10}{110} x$

$\frac{283}{383} x$

Q. 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:

x

283383​x

11010​x

383283​x

Let the mass of the gas be m. At a fixed temperature and pressure, volume is fixed. Density of the gas, ρ=Vm​ Now Pρ​=PVm​=nRTm​ ⇒nRTm​=x (By question) ⇒xT=constant⇒x1​T1​=x2​T2​ ⇒x2​⇒T2​x1​T1​​=383283​x⎣⎡​∴T1​T2​​==​283K383K​⎦⎤​

Q. At $1 0^{\circ}$ C the value of the density of a fixed mass of an ideal gas divided by its pressure is $x$ . At $11 0^{\circ}$ C this ratio is:

$x$

$\frac{383}{283} x$

$\frac{10}{110} x$

$\frac{283}{383} x$