SATHEE: Thermodynamics Question 140

Thermodynamics Question 140

Question: In a thermodynamics process, pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8J of work is done on the gas. If the initial internal energy of the gas was 30J. The final internal energy will be

[DPMT 2002]

Options:

A) 18J

B) 9J

C) 4.5J

D) 36J

Show Answer

Answer:

Correct Answer: A

Solution:

Given $Δ Q = - 20 J,$

$Δ W = - 8 J$ and $U_{1} = 30 J$

$Δ Q = Δ U + Δ W$

Therefore $Δ U = (Δ Q - Δ W)$

Therefore $(U_{f} - U_{i})$ = $(U_{f} - 30) = - 20 - (- 8)$

Therefore $U_{f} = 18 J$

Thermodynamics Question 140

Question: In a thermodynamics process, pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8J of work is done on the gas. If the initial internal energy of the gas was 30J. The final internal energy will be

Options:

Answer:

Solution:

Engineering Preparation

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NCERT Books & Solutions

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Other Resources

Syllabus

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Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Thermodynamics Question 140

Question: In a thermodynamics process, pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8J of work is done on the gas. If the initial internal energy of the gas was 30J. The final internal energy will be

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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