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A cylindrical container is shown in figure-2.7 in which a gas is enclosed. Its initial volume is Vand temperature is T. As no external pressure is applied on the light piston shown, gas pressure must be equal to the atmospheric pressure. If gas temperature is doubled, find its final volume. In its final state if piston is clamped and temperature is again doubled, find the final pressure of the gas. |
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Answer» Solution :In the initial state the pressure, volume and temperature of GAS `P_(0)` and T respectively if `P_(0)` is the atmospheric pressure. It is given that temperature of gas is INCREASED to double its value i.e. up to 2T. As in initial and final state pressure of gas remains constant as initially as well as finally the PISTON is exposed to atmospheric pressure. Thus we have from charls and Gay Lussac LAW `(V_(1))/(T_(1))=(V_(2))/(T_(2))` Here `""V_(2)=(V_(1)T_(2))/(T_(1))=2V` `["As"V_(1)=V,T_(1)=T" and "T_(2)=2T]` Thus after increasing the gas temperature to 2T, its volume becomes 2V. Now the piston is clamped that means, now the volume of gas remains constant. Again its temperature isdoubled from 2T to 4r, let the pressure changes from `P_(0)` to P'. Thus we have `(P_(1))/(T_(1))=(P_(2))/(T_(2))` or `""P_(2)=P'=(P_(1)T_(2))/(T_(1))=2P_(0)` `["As"P_(1)=P_(0),T_(1)=2T" and "T_(2)=4T]` Thus in final state gas pressure becomes `2P_(0)`. |
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