The density of a gas at normal pressure and 27oC temperature is 24 units. Keeping the pressure constant, the density at 127oC in same units will be

Question

The density of a gas at normal pressure and 27oC temperature is 24 units. Keeping the pressure constant, the density at 127oC in same units will be (A) 6 (B) 12 (C) 18 (D) 24

Tardigrade Admin · Accepted Answer

Pressure is P=(ρ R T/M) Hence at constant pressure, ρ T= constant ⇒ (ρ 1/ρ 2)=(T1/T2) ⇒ (24/(ρ )2)=((273 + 127)/(273 + 27))=(400/300) ⇒ p2=18 units

Q. The density of a gas at normal pressure and $2 7^{o} C$ temperature is $24$ units. Keeping the pressure constant, the density at $12 7^{o} C$ in same units will be

$6$

$12$

$18$

$24$

Pressure is $P = \frac{ρRT}{M}$
Hence at constant pressure, $ρT =$ constant
$\Rightarrow \frac{ρ _{1}}{ρ _{2}} = \frac{T _{1}}{T _{2}}$
$\Rightarrow \frac{24}{( ρ ) _{2}} = \frac{( 273 + 127 )}{( 273 + 27 )} = \frac{400}{300}$
$\Rightarrow p_{2} = 18 u ni t s$

Q. The density of a gas at normal pressure and 27oC temperature is 24 units. Keeping the pressure constant, the density at 127oC in same units will be

6

12

18

24

Pressure is P=MρRT​ Hence at constant pressure, ρT= constant ⇒ρ2​ρ1​​=T2​T1​​ ⇒(ρ)2​24​=(273+27)(273+127)​=300400​ ⇒p2​=18units

Q. The density of a gas at normal pressure and $2 7^{o} C$ temperature is $24$ units. Keeping the pressure constant, the density at $12 7^{o} C$ in same units will be

$6$

$12$

$18$

$24$

Pressure is $P = \frac{ρRT}{M}$
Hence at constant pressure, $ρT =$ constant
$\Rightarrow \frac{ρ _{1}}{ρ _{2}} = \frac{T _{1}}{T _{2}}$
$\Rightarrow \frac{24}{( ρ ) _{2}} = \frac{( 273 + 127 )}{( 273 + 27 )} = \frac{400}{300}$
$\Rightarrow p_{2} = 18 u ni t s$