Question

One liter of a gas weighs 2g at 300k and 1atm pressure.If the pressure is made 75 atm,at which of the following temperature will 1L of the same gas weighs 1g.? Solve it ?

Answer 1

Answer:

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Answer 2

Answer:

At 450 K temperature will give 1L of the same gas weighs 1 gram.

Explanation:

According to the ideal gas equation ,  $PV = n RT$

P is the pressure of the gas

V is the volume of the gas

n is the number of moles of gas

R is the gas constant

T is the temperature of the gas in kelvin.

we can rewrite the ideal gas equation in terms of mass

that is,  $PV = \frac{m}{M} RT$

m is the given mass

M is the molar mass of gas

From the question,

P1 = 1 atm

V1 = 1 L

m1 = 2 g

T1 = 300 K

P2 = 0.75 atm(In question pressure written as 75  atm, pressure is too high pressure, that condition is not possible )

V2 = 1 L

m2 = 1 g

M1 = M2 = M

T2 =?

$\frac{P1V1}{P2V2} = \frac{\frac{m1}{M}RT1 }{\frac{m2}{M} RT2}$

$\frac{P1V1}{P2V2} = \frac{m1 T1}{m2 T2}$

T2 =  $\frac{ 2*300*0.75*1}{ 1*1*1} = 450 K$