Question

An ideal gas is in equilibrium at initial state with temperature T=137oC, pressure P = 0.75Pa and volume V = 0.75 m3. If there is a change in state in which the gas undergoes an isothermal process to a final state of equilibrium during which the volume is doubled. Calculate the temperature and pressure of the gas at this final state.

Answer 1

since the gas is in an equilibrium state therefore we can consider that Both Temp.& Pressure
hav constant values
so we can apply both Boyle's & Charle's law
Boyle's law: V\[\alpha\] 1/P At constant T
so the pressure will be halved
charle's law: V\[\alpha\] T At constant P
so the Temp. will be doubled Too

Answer 2

thanks but its not working. i tried \[V_{1}P_{1}\div T_{1} = V_{2}P_{2}\div T_{2}\] but its not working. How can i get both pressure and temperature at different times when they are unknown in the same equation.

Answer 3

well
u hav the volume
& the volume has an equation with pressure & other with Temp.
each equation is single with itself why u mixed them ?!!

Answer 4

Please express the equation i should use, if i can't use the ideal gas situation which is not required or better still can u solve it for me and show me the steps so i can learn?

Answer 5

kindly reply? i will be back online in some minutes back

Answer 6

\[v1\div v2=p2\div p1\]
&
\[v1\div v2=t1\div t2\]

Answer 7

Given that the reaction is isothermal we know that we have no variations in temperature, so T=T2=137C.Now when the pressure doubles we set P2=2P=1.5Pa.From boyle's law now we have P2*V2=P*V (the boyles law states that in an ideal gas the product of pressure and volume is constant) from the above we get V2=0.375Pa

Answer 8

the initial and final temperature is same since it is an isothermal process.. i.e. T1=T2 .. Now use P1/V1 = P2/V2

Answer 9

We can surely use the ideal gas law here.

First, let's fix our states as best we can with the information given in the problem.

1) Isothermal process: \(T_1 = T_2\).
2) Volume is doubled: \(V_1 = {1 \over 2}V_2\)
3) Assume a closed system: \(n_1 = n_2\).

Therefore, from the ideal gas law, we get\[{P_1 V_1 \over n_1 R T_1} = {P_2 V_2 \over n_2 R T_2}\]From #1 and #3, we can simplify the above expression to be\[{P_1 V_1 \over T_1} = {P_2 V_2 \over T_1} \rightarrow P_1 V_1 = P_2V_2\]From #2, we can further simplify it to be\[{P_1 V_1} = {P_2 (2V_1) }\]\[P_2 = {P_1 V_1 \over 2V_1} = {P_1 \over 2}\]The final expression is Boyle's Law, but it's nice to be able to work through it from the ideal gas law in-case we forget under what conditions Boyle's Law holds (isothermal and constant amount).

Answer 10

thanks so far