Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures . (a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T2? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If we obtained similar plots for 1.00×10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mo1¯¹ K¯¹.)

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Solution

(a).

The expression for ideal gas equation is,

PV=nRT(1)

Here, P is the pressure of the gas, R is the universal gas constant, V is the volume, T is the temperature and n is the number of moles.

From equation (1),

PV T =nR

For an ideal gas, the term nR will remain constant.

Hence, PV T is also constant. This is shown by the dotted line in the given graph.

(b).

When the temperature of a real gas is increased, it approaches to the behavior of an ideal gas.

In the given graph, gas which is at temperature T 1 is closer to the dotted line which is a line of an ideal gas showing an ideal gas behavior. As the T 2 gas line is closer to the dotted line, therefore its temperature must be greater than that of T 2 .

Therefore, T 1 > T 2 is true.

(c).

Given, the mass of oxygen gas is 10 −3 kg or 1 g and the molecular mass of oxygen is 32 g.

Let n be the number of moles, then

n= molar mass molecular mass = 1 32

Value of PV T from equation (1) is,

PV T =nR

Substitute the values in the above expression.

PV T = 1 32 ×8.314 =0.26 J/K

Hence, the value of PV T where curves meet on y-axis is 0.26 J/K .

(d).

As the value of nR is different for each gas, therefore the value of PV T will be different in the case of hydrogen.

Given, the value of PV T is same for hydrogen as that in oxygen, so

PV T =0.26 nR=0.26 m M R=0.26

Here, m is the mass and M is the molar mass.

Substitute the values in the above equation.

m M R=0.26 m= 0.26M R = 0.26×2.02 8.314 =6.3× 10 −5 kg

Hence, 6.3× 10 −5 kg of hydrogen will have the same value of PV T as that of oxygen.

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Similar questions

Q. Figure 13.8 shows plot of PV/T versus P for

1.00 \times 10^{- 3} k g

of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true :

T_{1} > T_{2} o r T_{1} < T_{2}

?
(c) What is the value of PV / T where the curves meet on the y-axis?
(d) If we obtained similar plots for

1.00 \times 10^{- 3} k g

of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of

H_{2} = 2.02 u

, of

O_{2} = 32.0 u

R = 8.31 J m o l^{- 1} K^{- 1}

Figure 13.8 shows plot of PV/T versus Pfor 1.00×10^–3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?

(b) Which is true: T₁> T₂ or T₁ < T₂?

(d) If we obtained similar plots for 1.00 ×10^–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H₂= 2.02 u, of O₂ = 32.0 u, R = 8.31 J mo1^–1 K^–1.)