An ideal gas increases in temperature by two different processes.... An ideal ga
ID: 2279528 • Letter: A
Question
An ideal gas increases in temperature by two different processes....
An ideal gas increases in temperature from 22 degree C to 42 degree C by two different processes. In one process, the temperature increases at constant volume, and in the other process the temperature increases at constant pressure. Which of the following statements about this gas are correct? (There may be more than one correct choice). Explain your choice(s). A) The heat required to cause this temperature change is the same for both the constant-volume and the constant-pressure processes. B) More heat is required for the constant-pressure process than for the constant-volume process. C) The change in the internal energy of the gas is the same for both the constant-volume and the constant-pressure processes. D) The root-mean-square (thermal) speed of the gas molecules increases more during the constant-volume process than during the constant-pressure process.Explanation / Answer
By the first law of thermodynamics, dQ=dW+dU, where dQis the heat supplied, dW the work done by the system and dU the change in internal energy.Now internal energy is a function of temperature so in both the cases its change will be same.Thus more heat will be required in the case in which more work is done by the system which is in case of constant pressure because work done by the system in constant volume case is 0.So, B and C are correct from the above explanation.Also, root mean square speed of gas molecules is a function of its temperature.So, increase in it will be same in both cases. Hence the answer is B and C.
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