(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℃.