Ideal gas heating

The above shows a resistor with resistance R R installed in a cylinder filled with a fixed quantity of ideal gas. We connect the resistor to a power supply with a voltage of V V to provide heating energy in the cylinder. Suppose the pressure and volume of the ideal gas before and after heating are ( P 0 , V 0 ) (P_0, V_0) and ( P 0 , 1.5 V 0 ) , (P_0, 1.5V_0), respectively. If the internal energy of the ideal gas before heating is U 0 , U_0, then what is the change in the amount of the internal energy Δ U \Delta U after heating? (The cylinder is sealed completely.)

1 2 U 0 \frac{1}{2}U_0 U 0 U_0 2 3 U 0 \frac{2}{3}U_0 3 2 U 0 \frac{3}{2}U_0

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2 solutions

Kaustav Mitra
Apr 13, 2014

Use the ideal gas law PV/T = constant. Assume the initial temperature to be 't' and final temperature to be 'T'. given the necessary information about initial P, V and final P, V, try to express T in terms of t. It comes out to be T= 1.5t, so change in temperature is 0.5t. If initial internal energy associated with temperature t is U, and since internal energy of an ideal gas is proportional to temperature thus change in internal energy energy is 0.5U or U/2.

Jaivir Singh
Apr 13, 2014

change in U is negative of W

change in U is negative of W, W is equals to Pressure*change in volume, change in volume= 0.5, so that change in internal energy is 1/2U

Ratnani Hiren - 7 years, 1 month ago

i think that in this case U is not negative of W since there is increment in U plus heat is supplied to cause increment in U and work done on the piston

Manish Bhandari - 7 years, 1 month ago

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