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 was

Options

A. 273 K
B. 200 K
C. 100 K
D. 300 K

Answer

2

Explanation

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