Ideal gas undergoing cycle change - find temperature

[Amith2006]

Sir,
One mole of an ideal gas undergoes a cycle change as shown in figure. The process A-B is isothermal. The pressure and volume at A is 1.1 x 10^5 N/m^2 and 22.4 litres respectively. What is the temperature of the gas at C?
If the temperature at one point was given the temperature at the other point could be calculated. Here only the pressure and volume are given.

 

[Andrew Mason]

Sir,
One mole of an ideal gas undergoes a cycle change as shown in figure. The process A-B is isothermal. The pressure and volume at A is 1.1 x 10^5 N/m^2 and 22.4 litres respectively. What is the temperature of the gas at C?
If the temperature at one point was given the temperature at the other point could be calculated. Here only the pressure and volume are given.

PV=nRT

AM

 

[kyle8921]

Thank you for your question. In order to find the temperature of the gas at point C in this cycle, we need to use the ideal gas law, which states that the pressure (P), volume (V), and temperature (T) of a gas are all related by the equation PV = nRT, where n is the number of moles of gas and R is the ideal gas constant.

Since we know the pressure and volume at point A, we can use this information to find the initial temperature (T1) of the gas at point A using the ideal gas law. This will serve as our starting point for the cycle.

T1 = (PV)/(nR) = (1.1 x 10^5 N/m^2)(22.4 L)/(1 mol)(8.314 J/mol*K) = 299.4 K

Now, since the process A-B is isothermal, we know that the temperature remains constant at T1 throughout this process. Therefore, the temperature at point B will also be 299.4 K.

Next, for the process B-C, we can use the fact that the volume of the gas remains constant at 22.4 L to find the final temperature (T2) at point C.

T2 = (PV)/(nR) = (0.55 x 10^5 N/m^2)(22.4 L)/(1 mol)(8.314 J/mol*K) = 149.7 K

Therefore, the temperature of the gas at point C is 149.7 K. I hope this helps to answer your question.