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0.1 mole of argon has pressure P & temperature TK in the vessel. On keeping the vessel at `50^@C` higher temperature, 0.8gm of argon was given out to maintain same pressure. The original temperature wasA. 273 KB. 200 KC. 100 KD. 300 K |
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Answer» Correct Answer - 2 0.1 x T=(0.1 x 0.02) x (T+50) 0.1 x T=0.08T+4 0.02 x T=4 T=200 K |
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