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**1. Homework Statement**

In an isolated chamber closed by a moving piston is an capsule with 0.5 l of air at 3 bar and 100 °C. Chamber is filled with 2 l of air at 1 bar and 20 °C. The same pressure and temperature surround the piston from the other side. Capsule breaks. What is the volume and temperature now? ##\kappa =1.4##

**2. Homework Equations**

##\Delta U= \Delta W + \Delta Q = \Delta W + 0## because of the isolation

##p_cV_c=\frac{m_c}{M}RT_c## - c for capsule

##p_0V_0=\frac{m_0}{M}RT_0## - 0 for gas before the capsule breaks

##p_0V=\frac{m_0+m_c}{M}RT## - no index for gas after the capsule breaks

**3. The Attempt at a Solution**

I would be really happy if somebody could check what am I doing wrong because I just can't get the same result as written in the book...

##\Delta U=m_cc_v(T-T_c)+m_0c_v(T-T_0)=(m_c+m_0)c_vT-m_cc_vT_c-m_0c_vT_0##

##\Delta W=-p_0(V-(V_c+V_0))##

##(m_c+m_0)c_vT-m_cc_vT_c-m_0c_vT_0=-p_0(V-(V_c+V_0))##

Now for all the indexes: ##m_iT_i=\frac{p_iV_iM}{R}## . Than:

##\frac{M}{R}c_v(p_0V-p_1V_1-p_0V_0)=-p_0V+p_0(V_c+V_0)##

for ## cv=\frac{R}{M(\kappa -1)}##

##\frac{1}{\kappa -1}(p_0V-p_1V_1-p_0V_0)+p_0V=p_0(V_c+V_0)##

##p_0V(\frac{1}{\kappa -1}+1)-\frac{1}{\kappa -1}(p_1V_1+p_0V_0)=p_0(V_c+V_0)##

##p_0V(\frac{\kappa }{\kappa -1}=p_0(V_c+V_0)+\frac{1}{\kappa -1}(p_1V_1+p_0V_0)##

##V=\frac{(\kappa -1)p_0(V_c+V_0)+(p_1V_1+p_0V_0)}{p_0\kappa }##

For data above, this gives me 3.2 l instead of 2.71 l.

http://web2.0calc.com/?q=(0.4*10**5*(0.0005+0.002)+3*10**5*0.0005+10**5*0.002)/(10**5*1.4

WHAT ON EARTH is wrong here? :/