This question can be solved using the Combined Gas Law, which relates pressure (PPP), volume (VVV), and temperature (TTT) for a given amount of gas:
P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}T1P1V1=T2P2V2
Given Data:
- P1=2 barP_1 = 2 \, \text{bar}P1=2bar
- V1=200 mlV_1 = 200 \, \text{ml}V1=200ml
- T1=100 KT_1 = 100 \, \text{K}T1=100K
- P2=5 barP_2 = 5 \, \text{bar}P2=5bar
- T2=200 KT_2 = 200 \, \text{K}T2=200K
- V2=?V_2 = ?V2=?
Substituting into the formula:
2⋅200100=5⋅V2200\frac{2 \cdot 200}{100} = \frac{5 \cdot V_2}{200}1002⋅200=2005⋅V2
Simplify:
400100=5V2200\frac{400}{100} = \frac{5V_2}{200}100400=2005V2 4=5V22004 = \frac{5V_2}{200}4=2005V2
Solve for V2V_2V2:
V2=4⋅2005=160 mlV_2 = \frac{4 \cdot 200}{5} = 160 \, \text{ml}V2=54⋅200=160ml
Final Answer:
(b) 160 ml