# The gas is enclosed in a metal container with movable piston on top . heat is added to the system from bottom by placing a candle at the bottom which of the following is true ?

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1)The temperature must go up if the piston remains stationary

2) The temperature must go up if the piston is pulled out dramatically

3)the temperature must go up no matter what happens to the piston .

4)The temperature must go down no matter what happens to the piston

1)The temperature must go up if the piston remains stationary

2) The temperature must go up if the piston is pulled out dramatically

3)the temperature must go up no matter what happens to the piston .

4)The temperature must go down no matter what happens to the piston

##### 1 Answer

The temperature must go up if the piston remains stationary.

#### Explanation:

**The temperature must go up if the piston remains stationary**

If the piston remains stationary, we know that

#DeltaV# , the change in volume of the system, is necessarily zero, i.e.#DeltaV=0# .This implies that the system does no work on the environment as it does not expand.

#W_"sys"=int" P dV"# where

#P# is the pressure of the system and#V# is the volumeBy the first law of thermodynamics:

#Q_"in"=W_"out"+DeltaE_"th"# If no work is done by the system,

#W_"out"=0# , so the first law statement becomes:

#=>Q_"in"=DeltaE_"th"# If there is no change in the temperature of the gas,

#Q# and#DeltaE# would be zero as well, which would not be possible for the situation.

#DeltaE=3/2Nk_bDeltaT# We can therefore convince ourselves that we must have a temperature change.

Assuming an ideal gas, we should be able to apply the ideal gas law:

#(PV)/T="constant"# i.e.

#(P_1V_1)/T_1=(P_2V_2)/T_2# Since

#V_1=V_2# , this simplifies to:

#P_1/T_1=P_2/T_2# For the ideal gas law to be valid, we must have that if the pressure decreases, the temperature decreases by the same amount, and vice versa.

It doesn't make much sense intuitively for the temperature of the gas to decrease, which would mean that the pressure also decreases, as long as we have the constant heat source and no work being done. This is also clear from our statement of

#Q_"in"# , which is a positive quantity as heat is being added to the system. Therefore#DeltaE# must also be a positive quantity. Since

#DeltaE=3/2Nk_bDeltaT#

#=>=3/2Nk_b(T_f-T_i)# and

#N# and#k_b# remain constant,#T_f>T_i# .

So we can conclude that the temperature *must increase*.