3000J of heat is given to a gas at a constant pressure of P=(2×105N/m2). If its volume increases by 10 litres during the process, find the change in the internal energy of the gas.
A
1600J
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B
3000J
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C
2500J
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D
1000J
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Solution
The correct option is D1000J Given: q=3000J P=2×105N/m2 △V=10×10−3m3 So, the work done is given as, w=−P△V=−(2×105N/m2)(10×10−3m3)=−2×103J
From the first law of thermodynamics, △U=q+w=3000−2000=1000J