
When 229J of energy is supplied as a heat at constant pressure to $3molAr\left( g \right)$, the temperature of the sample is increased by $2.55K$. Calculate the molar heat capacity at constant volume.
(1)$30KJ{K^{ - 1}}mo{l^{ - 1}}$
(2)$30J{K^{ - 1}}mo{l^{ - 1}}$
(3)$21.7J{K^{ - 1}}mo{l^{ - 1}}$
(4)$21.7KJ{K^{ - 1}}mo{l^{ - 1}}$
Answer
546k+ views
Hint: We know that the molar heat capability of a chemical substance is the amount of energy that has to be added, within the variety of heat, to at least one mole of the substance so as to cause a rise of 1 unit in its temperature. Alternatively, it's the warmth capacity of a sample of the substance divided by the number of substances of the sample; or additionally the particular heat capacity of the substance times its molar mass. The SI unit of heat is joule per Kelvin per mole.
Complete step by step solution:
Given data contains;
$\partial H = 229J$
$n = 3mol$
$\partial T = 2.55K$
We know that the molar heat capacity at constant pressure is given by,
${C_{p,m}}\, = \dfrac{{\partial H}}{{n\partial T}}$
Now we calculate the molar heat capacity at constant pressure by substituting the known values in the above equation.
${C_{p,m}}\, = \dfrac{{229J}}{{3mol \times 2.55K}} = 29.93J{K^{ - 1}}mo{l^{ - 1}}$
Molar heat capability at constant pressure and molar heat capacity at constant volume are related. So, currently we will calculate the molar heat capacity at constant volume by mistreatment the equation,
${C_p} - {C_v} = R$
Let we substitute the values in the above equation we get,
$ \Rightarrow 29.93J{K^{ - 1}} - {C_v} = 8.314J{K^{ - 1}}mo{l^{ - 1}}$
${C_v} = 29.93J{K^{ - 1}} - 8.314J{K^{ - 1}}mo{l^{ - 1}}$
On simplifying we get,
$ \Rightarrow {C_v} = 21.616J{K^{ - 1}}mo{l^{ - 1}}$
The obtained value is approximately equal to option (3).
Therefore,option (3) is the correct option.
Note:
We have to remember that the injection of warmth energy into a substance, besides raising its temperature, sometimes causes a rise in its volume and/or its pressure, counting on however the sample is confined. The selection created concerning the latter affects the measured molar heat capacity, even for identical beginning pressure P and starting temperature T. 2 specific selections are widely used. If the pressure is unbroken constant (for instance, at the close region pressure), and therefore the sample is allowed to expand, the growth generates work because the force from the pressure displaces the enclosure. that job should return from the warmth energy provided. The value so obtained is claimed to be the molar heat capability at constant pressure (or isobaric), and is usually denoted \[cP,m\] etc. On the opposite hand, if the growth is prevented to Illustrate by a sufficiently rigid enclosure, or by increasing the external pressure to counteract the inner one no work is generated, and therefore the heat that will have gone into it should instead contribute to the internal energy of the object, as well as raising its temperature by an additional amount. The worth obtained in this manner is said to be the molar heat capacity at constant volume (or isochoric) and denoted \[cV,m\] etc. The value of \[cV,m\] is typically below the worth of \[cP,m\]. This distinction is especially notable in gases wherever values under constant pressure are typically \[30\% \] to \[66.7\% \] larger than those at constant volume. All strategies for the activity of heat apply to molar heat capability as well.
Complete step by step solution:
Given data contains;
$\partial H = 229J$
$n = 3mol$
$\partial T = 2.55K$
We know that the molar heat capacity at constant pressure is given by,
${C_{p,m}}\, = \dfrac{{\partial H}}{{n\partial T}}$
Now we calculate the molar heat capacity at constant pressure by substituting the known values in the above equation.
${C_{p,m}}\, = \dfrac{{229J}}{{3mol \times 2.55K}} = 29.93J{K^{ - 1}}mo{l^{ - 1}}$
Molar heat capability at constant pressure and molar heat capacity at constant volume are related. So, currently we will calculate the molar heat capacity at constant volume by mistreatment the equation,
${C_p} - {C_v} = R$
Let we substitute the values in the above equation we get,
$ \Rightarrow 29.93J{K^{ - 1}} - {C_v} = 8.314J{K^{ - 1}}mo{l^{ - 1}}$
${C_v} = 29.93J{K^{ - 1}} - 8.314J{K^{ - 1}}mo{l^{ - 1}}$
On simplifying we get,
$ \Rightarrow {C_v} = 21.616J{K^{ - 1}}mo{l^{ - 1}}$
The obtained value is approximately equal to option (3).
Therefore,option (3) is the correct option.
Note:
We have to remember that the injection of warmth energy into a substance, besides raising its temperature, sometimes causes a rise in its volume and/or its pressure, counting on however the sample is confined. The selection created concerning the latter affects the measured molar heat capacity, even for identical beginning pressure P and starting temperature T. 2 specific selections are widely used. If the pressure is unbroken constant (for instance, at the close region pressure), and therefore the sample is allowed to expand, the growth generates work because the force from the pressure displaces the enclosure. that job should return from the warmth energy provided. The value so obtained is claimed to be the molar heat capability at constant pressure (or isobaric), and is usually denoted \[cP,m\] etc. On the opposite hand, if the growth is prevented to Illustrate by a sufficiently rigid enclosure, or by increasing the external pressure to counteract the inner one no work is generated, and therefore the heat that will have gone into it should instead contribute to the internal energy of the object, as well as raising its temperature by an additional amount. The worth obtained in this manner is said to be the molar heat capacity at constant volume (or isochoric) and denoted \[cV,m\] etc. The value of \[cV,m\] is typically below the worth of \[cP,m\]. This distinction is especially notable in gases wherever values under constant pressure are typically \[30\% \] to \[66.7\% \] larger than those at constant volume. All strategies for the activity of heat apply to molar heat capability as well.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

How many 5 digit telephone numbers can be constructed class 11 maths CBSE

Draw a well labelled diagram of reflex arc and explain class 11 biology CBSE

What is the difference between noise and music Can class 11 physics CBSE

Trending doubts
In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

10 examples of friction in our daily life

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

