(1) Taking the whole as the research object, the force analysis is carried out: the total gravity mAg+mBg, the ground supporting force N 1' and the atmospheric pressure offset. According to the equilibrium conditions, it is known that N1'= MAG+MBG = (10+20) × 6544. N=300N, which is obtained from Newton's third law: the pressure of B on the ground n1= n1'= mag+mbg = (10+20) ×10n = 300 n.

(2) In the same way, the pressure N2 = MAG +MBG = (10+20) × 10 was obtained. N=300N？

When the temperature of gas in B is t 1=27℃,

Taking piston A as the research object, we can get p 1S+mAg=p0S, p1= p0-mags = (105-10×1050×10? 4)? Pa=0.8× 105Pa，T 1=(273+27)K=300K

When the gas temperature in B is t2,

Taking cylinder B as the research object, according to the equilibrium conditions p0S+mBg=p2S, P2 = P0+MBGS = (105+20×1050×108? 4)? Pa= 1.4× 105Pa

According to Charles theorem

p 1T 1=p2T2

T2 = P2P 1t 1 = 1.4× 1050.8× 105×300k = 525k。

Therefore, t2=T2-273=252℃

Answer: (1)B Ground pressure, N 1, is 300 n;

(2) The pressure N2 of A on the ground is 300N, and the gas temperature t2 is 252℃.