Question

at the top of mountain, the thermometer reads 0degree and the barrow meter reads 710mmhg .at the bottom of the mountain the temperature is30degrees and the pressure is 760mmhg.the ratio of the density of air at the top with the bottom is

Answer 1

we know, density = mass/volume
e.g., density is inversely proportional to volume .
∴ $\bold{\frac{d_{bottom}}{d_{top}}=\frac{v_{top}}{v_{bottom}}}$
We also know,
PV = nRT ,
so, V is directly proportional to (T/P)
Hence, $\bold{\frac{d_{bottom}}{d_{top}}=\frac{T_tP_b}{T_bP_t}}$
Given, Tt is the temperature at the top , Tb is the temperature at the bottom ,Pt is the pressure at the top and Pb is the pressure at the bottom.
Now,
$\bold{\frac{d_{bottom}}{d_{top}}}$ = (273 + 0)K × 760mm/(273+30)K × 710mm
= 273 × 76/303 × 71
= 0.9644

Answer 2

Explanation:

Acc-to Ideal gas equation:

d=RM/RT

d1=P1M/RT1 d2=P2M/RT2

d1/d2=P1M/RT1 / P2M/RT2 P1=760 mm=1 atm

T1=30 degree Celsiu d1/d2=P1/T1 / P2/T2 P2=710mm=0.934

d1/d2=P1 into T2 P2 into T1 T2=0 degree Celsiu=273k

d1/d2=1 into 300 / 0.934

into 273

d1/d2=1.18

d1=1.18 into d2...............