The answer is B, lithium would become positive and bromium would become negative.
I can't see the screw (if you'll pardon the expression), so I don't know a thing about it, and can't answer the question. There's no given information in the question.
Answer:
Approximately 
, if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.
Explanation:
Let 
and 
denote the pressure of this gas before and after the changes.
Let 
and 
denote the volume of this gas before and after the changes.
Let 
and 
denote the temperature (in degrees Kelvins) of this gas before and after the changes.
Let 
and 
denote the quantity (number of moles of gas particles) in this gas before and after the changes.
Assume that this gas is an ideal gas. By the ideal gas law, the ratios 
and 
should both be equal to the ideal gas constant, 
.
In other words:
.
.
Combine the two equations (equate the right-hand side) to obtain:
.
Rearrange this equation for an expression for , the temperature of this gas after the changes:
.
Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: 
, 
.
![\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20T_2%20%26%3D%20%5Cfrac%7BP_2%7D%7BP_1%7D%20%5Ccdot%20%5Cfrac%7BV_2%7D%7BV_1%7D%20%5Ccdot%20%5Cfrac%7Bn_1%7D%7Bn_2%7D%5Ccdot%20T_1%20%5C%5C%5B0.5em%5D%20%26%3D%20%5Cfrac%7B1.67%5C%3B%20%5Crm%20atm%7D%7B1.12%5C%3B%20%5Crm%20atm%7D%20%5Ctimes%20%5Cfrac%7B11.2%5C%3B%20%5Crm%20m%5E%7B3%7D%7D%7B15.7%5C%3B%20%5Crm%20m%5E%7B3%7D%7D%20%5Ctimes%201%20%5Ctimes%20245%5C%3B%20%5Crm%20K%20%5C%5C%5B0.5em%5D%20%26%5Capprox%20261%5C%3B%20%5Crm%20K%5Cend%7Baligned%7D)
.
The answer is A. Hope this helps I tried my best!
It's A they repel eachother
