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# A monatomic ideal gas expands slowly to twice its original volume, doing 450 J of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is (a) isothermal; (b) adiabatic; (c) isobaric.

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## Given

we are given a monatomic ideal gas that expands to twice its original volume which means $V_{2} = 2 V_{1}$ with work done $W$= 450 J. We are asked to calculate the heat added $(Q)$ to the gas and the change in internal energy $\Delta U$ if the process be (a) Isothermal (b) Adiabatic (c) Isobaric

## Known

We all know that for a gas doing a work by heat or even insulated, the first law of thermodynamics is applied in this case which relates the heat, the work done and the change in internal energy and would be used to calculate $Q$ and $\Delta U$ in this case.

$Q = W + \Delta U$

## Solution

(a) For the isothermal process, the temperature is constant during the process which means the change in temperature $\Delta T$ = 0. As the change in internal energy depends directly on $\Delta T$, we could obtain $\Delta U$ equals zero by

$\Delta U = nC_{V}\Delta T = 0$

Now plug the values for $W$ and $\Delta U$ into equation (1) to get $Q$ when the process is isothermal

\begin{aligned} Q &= W + \Delta U\\ &= 450 \,\text{J} + 0 \\ &= 450 \,\text{J} \end{aligned}

For the isothermal process all heat goes to increase the work done. $\boxed{Q = 450 \,\text{J}}$ and $\boxed{\Delta U = 0}$

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