Question

#CHEMISTRY

Q. At 100°C, the ratio of density of a fixed mass of an ideal gas to its pressure is X. What will be this ratio at 174.6°C?

ANSWER THIS QUESTION WITH PROPER EXPLANATION

#No Spamming Plz​

Answer 1

Answer:

P=dRT/M

P1/d1T1=P2/d2T2

Now,d1/P1=X

So,1/XT1=P2/d2T2

So,d2/P2=XT1/T2

d2/P2=X*(100+273)/(174.6+273)

=X*373/447.6

=0.83X.

Answer 2

Solution :

⏭ Given:

✏ At 100°C, the ratio of density of a fixed mass of an ideal gas to its pressure is X.

⏭ To Find:

✏ Ratio of density to the pressure at 174.6°C.

⏭ Conversation:

✒ 0°C = 273K

✒ T1 = 100°C = (100 + 273) = 373K

✒ T2 = 174.6°C = (174.6 + 273) = 447.6K

⏭ Formula:

✏ Formula of pressure exerts by ideal gas in terms of density of gas, gas constant, temperature of gas and molecular weight of gas is given by...

$\bigstar \: \boxed{ \tt{ \large{ \pink{P = \frac{dRT}{Mw}}}}} \: \bigstar$

⏭ Terms indication:

• P denotes pressure
• d denotes density
• R denotes gas constant
• T denotes temperature
• Mw denotes molar masd

⏭ Calculation:

✏ R and Mw have constant values in this question...

$\therefore \sf \: \red{ \dfrac{dT}{P} = constant} \\ \\ \implies \sf \: \frac{d_1T_1}{P_1} = \frac{d_2T_2}{P_2} \\ \\ \green{ \dag \sf \: putting \: all \: values, \: we \: get...} \\ \\ \implies \sf \: (x ) \times 373 = \dfrac{d_2}{P_2} \times 447.6 \\ \\ \implies \sf \: \dfrac{d_2}{P_2} = \frac{373x}{447.6} \\ \\ \implies \boxed{ \tt{ \orange{ \large{d_2 :P_2 = 5x : 6 }}}} \: \purple{ \bigstar}$