Question

Газ в количестве 1 моль совершает цикл, состоящий из 2 изохор и 2 изобар. наименьший объем газа 10л, наибольший-20л. наименьшее давление 2,5атм, наибольшее - 5 атм. наидите работу за цикл

Answer 1

и последнее:

А41=рmin(vmin-vmax);

A=Pmax(Vmax-Vmin)-Pmin(Vmin-Vmax)=(Vmax-Vmin)(Pmax-Pmin)=(2*10^-2(степень)м3(в кубе)-10^-2м3)(5*10^5Па-2,5*10^5Па)=2,5*10^3Дж

 

 

 

 

 

 

 

 

Answer 2

Calculation of Work for the Gas Cycle

To calculate the work done by the gas during the cycle, we can use the formula:

Work = -PΔV

where P is the pressure and ΔV is the change in volume.

In this case, the gas undergoes a cycle consisting of 2 isochoric (constant volume) processes and 2 isobaric (constant pressure) processes. The smallest volume of the gas is 10 liters, and the largest volume is 20 liters. The smallest pressure is 2.5 atm, and the largest pressure is 5 atm.

Let's calculate the work done during each process and then sum them up to find the total work done for the cycle.

Isochoric Processes

During an isochoric process, the volume remains constant, so the work done is zero.

Isobaric Processes

During an isobaric process, the pressure remains constant, so the work done can be calculated using the formula mentioned earlier.

Process 1: - Volume change (ΔV) = 20 L - 10 L = 10 L - Pressure (P) = 2.5 atm - Work (W) = -PΔV = -(2.5 atm)(10 L) = -25 atm·L

Process 2: - Volume change (ΔV) = 10 L - 20 L = -10 L (negative because the volume decreases) - Pressure (P) = 5 atm - Work (W) = -PΔV = -(5 atm)(-10 L) = 50 atm·L

Total Work for the Cycle

To find the total work done for the cycle, we sum up the work done during each process.

Total work = Work (Process 1) + Work (Process 2) + Work (Isochoric Processes)

Total work = -25 atm·L + 50 atm·L + 0 atm·L = 25 atm·L

Therefore, the work done by the gas during the cycle is 25 atm·L.