Thermodynamics

Classical Mechanics Level 3

Two moles of an ideal gas at temperature 300 K 300 \text{ K} was cooled isochorically so that the pressure was reduced to half. Then in an isobaric process the gas expanded till the temperature got back to initial value.

Find the amount of heat absorbed by the gas in the entire process.

Consider the value of gas constant to be 8.3 J mol × K 8.3 \dfrac { \text J }{ \text{mol}\times \text K } .

3167 J 3167 \text{ J} 2490 J 2490 \text{ J} 4638 J 4638 \text{ J} 2960 J 2960 \text{ J} 3451 J 3451 \text{ J} 342 J 342 \text{ J} 9234 J 9234 \text{ J} 3126 J 3126 \text{ J}

Jaswinder Singh by Jaswinder Singh , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Jaswinder Singh
Feb 21, 2016

For the isochoric process, since volume is constant Pressure ∝ Temperature . \text{ Pressure } \propto \text{ Temperature} . Final temperature will be half of the initial temperature: Δ T = − 300 2 \Delta T=\frac { -300 }{ 2 } Δ Q = n C V Δ T = − 300 C V \Delta Q=n{ C }_{ V }\Delta T=-300{ C }_{ V } For isobaric process: Δ T = 300 − 300 2 = 300 2 \Delta T=300-\frac { 300 }{ 2 } =\frac { 300 }{ 2 } Δ Q = n C P Δ T = 2 × 300 2 × ( C V + R ) = 300 C V + 300 R \Delta Q=n{ C }_{ P }\Delta T=2\times \frac { 300 }{ 2 } \times ({ C }_{ V }+R)=300{ C }_{ V }+300R

Total: Δ Q total = − 300 C V + 300 C V + 300 R = 300 R . \Delta { Q }_{ \text{total}}=-300{ C }_{ V }+300{ C }_{ V }+300R=300R.

8 Helpful 0 Interesting 0 Brilliant 0 Confused

In the second process, why gas has not done work=PdV due to change of volume?

Anandhu Raj - 5 years, 2 months ago

Log in to reply

No value for any value is provided ... You cannot assume any value .. And pdV=nRdT you can use both the equation but at the same time you have to consider the information provided as well... Moreover you have been asked for Q not the work done

Jaswinder Singh - 5 years, 2 months ago

Let T 1 = 300 k T_{1} = 300k

For an ideal gas at constant volume,
P α T P \alpha T
T 2 = T 1 P 2 T 2 = 150 K T_{2} = \dfrac{T_{1}P_{2}}{T_{2}} = 150 K
d T = 150 − 300 = − 150 K dT = 150-300 = -150 K


Using first law of thermodynamics ,
Δ Q = d U + Δ W \Delta Q = dU + \Delta W
During the isochoric process, Δ W = 0 \Delta W = 0 & d U = n C v d T dU = nC_{v}dT .
Δ Q 1 = n C v d T + 0 \Delta Q_{1} = nC_{v}dT + 0
During the isobaric process, the gas is brought to the initial state.
d T 2 = 300 − 150 = 150 K = − d T dT_{2} = 300-150 = 150K = -dT
Δ Q 2 = − n C v d T + P Δ V \Delta Q_{2} = -nC_{v}dT + P\Delta V
Δ Q = Δ Q 1 + Δ Q 2 \Delta Q = \Delta Q_{1} + \Delta Q_{2} .
Δ Q = n C v d T − n C v d T + P Δ V \Delta Q = nC_{v}dT - nC_{v}dT + P\Delta V
∴ Δ Q = P Δ V \therefore \Delta Q = P\Delta V
For an ideal gas,
P Δ V = n R Δ T = n R d T 2 P\Delta V = nR\Delta T = nRdT_{2}
∴ Δ Q = n R ( 300 − 150 ) = 2 × 8.3 × 150 = 2490 J \therefore \Delta Q = nR(300-150) = 2 \times 8.3 \times 150 = 2490 J

5 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...