An ideal gas increases in temperature from 22°C to 42°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 process requires more heat or are the required amount of heat same in both?

The process that requires more heat is the constant-pressure process than the constant-volume process.

According to the first law of thermodynamics, the heat that's applied to the system will be the addition of the change in internal energy and the work done.

In a constant-volume process, the work done on the gas is equal to zero. More heat will be required for the constant-pressure process than for the constant-volume process.

Also, it should be noted that the change in the thermal energy of the gas will be the same for the constant-pressure process and the constant-volume process.

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a- More heat is required for the constant-pressure process than for the constant-volume

Explanation:

we have to solve using the thermodynamic first law. this is the heat applied to the system

dQ = dU + dW

definition of terms:

dU = change in internal energy

dW = work done

we have it that

change in internal energy dU is directly proportional to work done dW

but when we are in constant volume process, work done of the gas is zero

therefore

dQ of constant pressure is > than that of constant volume

so constant pressure process requires more heat

hey , the answer is c

it is a negative acceleration