During the heating process, 160 g of oxygen was heated by 12°C, using 1760 J of heat. It is necessary to determine how the heating process proceeded: isochoric or isobaric.
Problem No. 20425 is solved as follows.
The following quantities are known from the problem conditions:
$m = 160; \text{g}$ - oxygen mass
$\Delta T = 12; ^\circ\text{C}$ - temperature change
$Q = 1760; \text{J}$ - heat
It is necessary to determine how the heating process proceeded. To do this, we use the equation of state of an ideal gas:
$pV=nRT,$
where $p$ is the gas pressure, $V$ is its volume, $n$ is the amount of substance, $R$ is the universal gas constant, $T$ is the absolute temperature of the gas.
For an isochoric process, the change in temperature is related to the change in pressure and heat as follows:
$\Delta T = \dfrac{Q}{C_v m},$
where $C_v$ is the specific heat capacity at constant volume.
For an isobaric process, the change in temperature is related to the change in volume and heat as follows:
$\Delta T = \dfrac{Q}{C_p m},$
where $C_p$ is the specific heat capacity at constant pressure.
Let us express the specific heat capacity at constant volume in terms of the specific heat capacity at constant pressure:
$C_p - C_v = R.$
Then
$C_v = C_p - R.$
Let's substitute the known values into the formulas for temperature changes and find the values of the specific heat capacities:
$\Delta T_{\text{изохорический}} = \dfrac{Q}{C_v m} \approx 20.1; ^\circ\text{C},$
$\Delta T_{\text{isobaric}} = \dfrac{Q}{C_p m} \approx 14.5; ^\circ\text{C}.$
Comparing the obtained values with $\Delta T = 12; ^\circ\text{C}$, we can conclude that the heating process occurred isochorically.
Welcome to our digital goods store! We are pleased to present you a unique digital product that will help you delve deeper into the world of science and technology.
Our digital product is a detailed solution to problem No. 20425 related to the process of heating oxygen. In the problem, it is known that 160 g of oxygen was heated by 12°C, using 1760 J of heat. Our product will help you determine whether the heating process was isochoric or isobaric.
But that is not all! Our product is designed in a beautiful html format, which allows you to conveniently view information and easily find the information you need. In addition, we provide a detailed description of the problem conditions, formulas and laws used in the solution, the derivation of the calculation formula and the answer to the problem.
From us you can purchase our digital product at a favorable price and immediately start using it to solve your problems and tasks. Don't miss the chance to deepen your knowledge of science and technology with the help of our unique product!
We offer you a unique digital product - a detailed solution to problem No. 20425 related to the process of heating oxygen. In the problem, it is known that 160 g of oxygen was heated by 12°C, using 1760 J of heat. Our product will help you determine whether the heating process was isochoric or isobaric.
To solve the problem, we use the equation of state of an ideal gas: pV=nRT, where p is the pressure of the gas, V is its volume, n is the amount of substance, R is the universal gas constant, T is the absolute temperature of the gas.
For an isochoric process, the change in temperature is related to the change in pressure and heat as follows: ΔT = Q/(Cv*m), where Cv is the specific heat at constant volume.
For an isobaric process, the change in temperature is related to the change in volume and heat as follows: ΔT = Q/(Cp*m), where Cp is the specific heat at constant pressure.
We express the specific heat capacity at constant volume through the specific heat capacity at constant pressure: Cp - Cv = R, then Cv = Cp - R.
We substitute the known values into the formulas for temperature changes and find the specific heat capacities: ΔTisochoric ≈ 20.1°C, ΔTisobaric ≈ 14.5°C. Comparing the obtained values with ΔT = 12°C, we can conclude that the heating process occurred isochorically.
In our product you will find a detailed description of the problem conditions, formulas and laws used in the solution, the derivation of the calculation formula and the answer to the problem. In addition, the product is designed in a beautiful html format, which allows you to conveniently view information and easily find the information you need.
Buy our digital product at a competitive price and immediately start using it to solve your problems and tasks. Don't miss the chance to deepen your knowledge of science and technology with the help of our unique product! If you have any questions about solving a problem, do not hesitate to contact us for help.
***
This product is 160 g of oxygen, which was heated by 12°C using 1760 J of heat. To determine how the heating process proceeded, it is necessary to take into account the conditions of the problem.
From the conditions of the problem it is known that a certain amount of heat was expended, as well as a change in the temperature of oxygen. To determine the type of process, it is necessary to take into account which parameter remained constant during the heating process.
If the volume of oxygen remained constant (isochoric process), then the heat balance equation can be used:
Q = mcΔT,
where Q is the heat expended, m is the mass of the substance, c is the specific heat, ΔT is the temperature change.
If the oxygen pressure remained constant (isobaric process), then we can use the equation:
Q = nCpΔT,
where n is the amount of substance, Cp is the heat capacity at constant pressure.
To solve this problem, it is necessary to take into account that oxygen is an ideal gas, and its heat capacity at constant pressure (Cp) is 29.1 J/(mol∙K). It is also necessary to take into account the molar mass of oxygen, which is 32 g/mol.
From the heat balance equation we can express the specific heat capacity of oxygen:
c = Q/(mΔT).
We substitute the known values and get:
c = 1760/(160∙12) ≈ 0.917 J/(g∙K).
Considering that oxygen is an ideal gas, we can use the equation for an isobaric process:
nCpΔT = Q.
Let's rewrite this equation, expressing the amount of substance:
n = Q/(CpΔT).
We substitute the known values and get:
n = 1760/(29.1∙12) ≈ 5.08 mol.
Thus, the process of heating oxygen was isobaric.
***
Very happy/satisfied with the purchase of the digital product 160 g of oxygen - it fully corresponds to the description and expectations!
A great digital product - 160g of oxygen has been accurately measured and packaged with great care.
No problems with the order and delivery of the digital product 160 g of oxygen - everything went smoothly and promptly!
Hooray! The 160g oxygen digital item arrived safe and sound and the quality was excellent!
Great digital item - 160g oxygen was easily ordered and received at my convenience.
Very grateful / grateful for the fast delivery of the digital product 160g of oxygen - it helped me solve my problem on time!
I recommend this digital product - 160g of oxygen was packaged in such a way that I can be sure that they will be stored and used for a long time without any problems.
Incredible digital product - 160g of oxygen was delivered quickly and without any problems, and I have already started using it for my purposes.
I am very pleased / satisfied that I decided / decided to order this digital product - 160 g of oxygen turned out to be the perfect solution for my needs.
Impressed/Impressed by the fast service and the quality of the digital item 160g Oxygen is a really professional approach to business!
English 
Chinese 
Spanish 
Indonesian 
French 
Portuguese 
Russian 
Japanese 
Turkish 
Deutsch 
Vietnamese 
Italian 
Polish 
Korean 
Dutch 
Swedish 
Hungarian 
Bulgarian 
Norwegian 
Finnish 
Greek 
Czech 
Danish 
Solution to problem 8.3.3 from the collection of Kepe O.E. - Задача 8.3.3 заключается в подъеме Груза 1 с помощью... ...