MHT CET · Chemistry · Thermodynamics (C)
If 100 L gas is enclosed in a cylinder, absorbs 302.6 J of heat and expands to 200 L against constant external pressure of 2 pascal. Calculate internal energy change of the gas.
- A 200 J
- B 300 J
- C 400 J
- D 100 J
Answer & Solution
Correct Answer
(D) 100 J
Step-by-step Solution
Detailed explanation
Given:
- \(q=302.6 \mathrm{~J}\),
- External pressure: \(P_{\text {ext }}=2 \mathrm{~pascal}\),
- Initial volume: \(V_1=100 \mathrm{~L}\),
- Final volume: \(V_2=200 \mathrm{~L}\),
- Work done: \(W=-P_{\mathrm{ext}} \Delta V\).
Step 1: Calculate Work Done (\(W\)):
\(W=-P_{\text {ext }}\left(V_2-V_1\right)=-(2)(200-100)=\) \(-200 \mathrm{~J}\)
Step 2: Use First Law of Thermodynamics:
\(\Delta U=q+W=302.6-200=-102.6 \mathrm{~J} .\)
Answer: 100 J, Option 4.
- \(q=302.6 \mathrm{~J}\),
- External pressure: \(P_{\text {ext }}=2 \mathrm{~pascal}\),
- Initial volume: \(V_1=100 \mathrm{~L}\),
- Final volume: \(V_2=200 \mathrm{~L}\),
- Work done: \(W=-P_{\mathrm{ext}} \Delta V\).
Step 1: Calculate Work Done (\(W\)):
\(W=-P_{\text {ext }}\left(V_2-V_1\right)=-(2)(200-100)=\) \(-200 \mathrm{~J}\)
Step 2: Use First Law of Thermodynamics:
\(\Delta U=q+W=302.6-200=-102.6 \mathrm{~J} .\)
Answer: 100 J, Option 4.
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