Question

The intercept on y-axis and slope of curve plotted between P/T vs. T
For an ideal gas having 10 moles in a closed rigid container of volume 8.21 L (P= Pressure in
atm and T=Temp. in K, log10 2= 0.30) are respectively.​

Answer 1

Answer:

The slope of the curve plotted between $\frac{P}{T}$ vs T is $0$ and the intercept on the y-axis of the curve is $0.1\ atm\ K^{-1}$·

Explanation:

Given that,

The number of moles of an ideal gas $=$ $10$ mol

The volume of the closed container $=$ $8.21$ L

The value of the log $10^{2}$ $=$ $0.30$

We have to find the slope and the intercept on the y-axis of the curve plotted between $\frac{P}{T}$ vs T·

It is mentioned in the question that, the container is a closed one· So here, the volume is constant·

At constant volume, an ideal gas obeys Gay-Lusacc's law· It is stated that "at constant volume, the pressure P of a fixed amount of gas is directly proportional to the absolute temperature T"·

That is,

 P ∝ T

 $\frac{P}{T} \ =$ a constant

To find out the slope, we know that

For an ideal gas,

 PV  $=$ nRT

 $\frac{P}{T} \ =$  $\frac{n\ R}{V}$

Here, the term  $\frac{P}{T}$  is independent of the term temperature·

On modifying this equation with the temperature term we get,

 $\frac{P}{T} \ =$ $(0)\ T\ +\frac{n\ R}{V}$

On comparing this equation with the straight-line equation,

y $=$ mx $+$ c

we get,

The slope m is zero·

Therefore,

The slope of the curve plotted between  $\frac{P}{T}$ vs T is  $0$·

To find out the intercept, we use the same modified ideal gas equation which is,

 $\frac{P}{T} \ =$ $0\ T\ +\frac{n\ R}{V}$

On comparing this equation with the y $=$ mx $+$ c equation,

we get,

The intercept term, c as  $\frac{n\ R}{V}$·

where,

n $=$ number of moles  $=$ $10$ mol

R  $=$ universal gas constant  $=$ $0.821\ L\ atm\ K^{-1}\ mol^{-1}$

V   $=$ Volume  $=$ $8.21$ L

On substituting the values we get,

Intercept, c  $=$ $\frac{10\ *\ 0.821}{8.21}$

Intercept, c  $=$ $0.1\ atm\ K^{-1}$

Therefore,

The intercept of the curve plotted between  $\frac{P}{T}$ vs T is  $0.1\ atm\ K^{-1}$·