1)
Дано: V=60 м^3, T=20 С=293К, T(р)=10 C=283К
Найти: m-?, φ-?
Решение:
![pV=\frac{m}{M}RT\\ p=\frac{mRT}{MV}\\ \frac{p}{m}=\frac{RT}{MV}\\ \frac{p}{m}=\frac{8,31*293}{(2+16)*10^{-3}*60}=2,25*10^3\\ m=\frac{p}{2,25*10^3}\\\\ T_p=\frac{b\gamma}{a-\gamma}\\ T_p(a-\gamma)=b\gamma\\ T_pa-T_p\gamma=b\gamma\\ T_pa=\gamma(T_p+b)\\ \gamma=\frac{T_pa}{T+b}\\ a=290K; \ b=510K;\\ \gamma=\frac{283*290}{283+510}=103\\ \gamma=\frac{aT}{T+b}+ln\phi\\ ln\phi=\gamma-\frac{aT}{T+b}=103-\frac{290*293}{510+293}=-2,8\\ \phi=\frac{1}{e^{2,8}}=\frac{1}{16,4}=0,06 \phi=\frac{p}{p_0}\\](https://tex.z-dn.net/?f=pV%3D%5Cfrac%7Bm%7D%7BM%7DRT%5C%5C+p%3D%5Cfrac%7BmRT%7D%7BMV%7D%5C%5C+%5Cfrac%7Bp%7D%7Bm%7D%3D%5Cfrac%7BRT%7D%7BMV%7D%5C%5C+%5Cfrac%7Bp%7D%7Bm%7D%3D%5Cfrac%7B8%2C31%2A293%7D%7B%282%2B16%29%2A10%5E%7B-3%7D%2A60%7D%3D2%2C25%2A10%5E3%5C%5C+m%3D%5Cfrac%7Bp%7D%7B2%2C25%2A10%5E3%7D%5C%5C%5C%5C+T_p%3D%5Cfrac%7Bb%5Cgamma%7D%7Ba-%5Cgamma%7D%5C%5C+T_p%28a-%5Cgamma%29%3Db%5Cgamma%5C%5C+T_pa-T_p%5Cgamma%3Db%5Cgamma%5C%5C+T_pa%3D%5Cgamma%28T_p%2Bb%29%5C%5C+%5Cgamma%3D%5Cfrac%7BT_pa%7D%7BT%2Bb%7D%5C%5C+a%3D290K%3B+%5C+b%3D510K%3B%5C%5C+%5Cgamma%3D%5Cfrac%7B283%2A290%7D%7B283%2B510%7D%3D103%5C%5C+%5Cgamma%3D%5Cfrac%7BaT%7D%7BT%2Bb%7D%2Bln%5Cphi%5C%5C+ln%5Cphi%3D%5Cgamma-%5Cfrac%7BaT%7D%7BT%2Bb%7D%3D103-%5Cfrac%7B290%2A293%7D%7B510%2B293%7D%3D-2%2C8%5C%5C+%5Cphi%3D%5Cfrac%7B1%7D%7Be%5E%7B2%2C8%7D%7D%3D%5Cfrac%7B1%7D%7B16%2C4%7D%3D0%2C06+%5Cphi%3D%5Cfrac%7Bp%7D%7Bp_0%7D%5C%5C+)
![p=\phi*p_0=0,06*2,33*10^3=0,14*10^3=140\\ m=\frac{p}{2,25*10^3}=\frac{140}{2,25*10^3}=62,2*10^{-3}=6,22*10^{-2}](https://tex.z-dn.net/?f=p%3D%5Cphi%2Ap_0%3D0%2C06%2A2%2C33%2A10%5E3%3D0%2C14%2A10%5E3%3D140%5C%5C+m%3D%5Cfrac%7Bp%7D%7B2%2C25%2A10%5E3%7D%3D%5Cfrac%7B140%7D%7B2%2C25%2A10%5E3%7D%3D62%2C2%2A10%5E%7B-3%7D%3D6%2C22%2A10%5E%7B-2%7D)
Ответ: 6,22*10^-2 кг, φ=0,06=6%
2)
Дано: V=75л=0,075м^3, p1=2,4*10^6 Па, T1=293К, p2=4*10^5 Па, T2=283К
Найти: Δm-?
Решение:
В данном масса непостоянна, т.к. горелка использует кислород для горения пропана.
При объем - константа, емкость горелки не меняется, записываем два уравнения и приравниваем их через константы(объем, газовоя постоянная, молярная масса)
![p_1V=\frac{m_1}{M}RT_1\\ V=\frac{m_1RT_1}{Mp_1}\\ \frac{MV}{R}=\frac{m_1T_1}{p_1}\\ \frac{MV}{R}=\frac{m_2T_2}{p_2}\\ \frac{m_1T_1}{p_1}=\frac{m_2T_2}{p_2}\\ \frac{m_1}{m_2}=\frac{T_2p_1}{p_2T_1}=\frac{283*2,4*10^6}{293*4*10^5}=5,8\\ V=\frac{m_1RT_1}{Mp_1}\\ m_1=\frac{VMp_1}{RT_1}=\frac{0,075*16*10^{-3}*2,4*10^6}{8,31*293}=1,2\\ \frac{m_1}{m_2}=5,8\\ m_2=\frac{m_1}{5,8}=\frac{1,2}{5,8}=0,2\\ \Delta m =m_1-m_2=1,2-0,2=1](https://tex.z-dn.net/?f=p_1V%3D%5Cfrac%7Bm_1%7D%7BM%7DRT_1%5C%5C+V%3D%5Cfrac%7Bm_1RT_1%7D%7BMp_1%7D%5C%5C+%5Cfrac%7BMV%7D%7BR%7D%3D%5Cfrac%7Bm_1T_1%7D%7Bp_1%7D%5C%5C+%5Cfrac%7BMV%7D%7BR%7D%3D%5Cfrac%7Bm_2T_2%7D%7Bp_2%7D%5C%5C+%5Cfrac%7Bm_1T_1%7D%7Bp_1%7D%3D%5Cfrac%7Bm_2T_2%7D%7Bp_2%7D%5C%5C+%5Cfrac%7Bm_1%7D%7Bm_2%7D%3D%5Cfrac%7BT_2p_1%7D%7Bp_2T_1%7D%3D%5Cfrac%7B283%2A2%2C4%2A10%5E6%7D%7B293%2A4%2A10%5E5%7D%3D5%2C8%5C%5C+V%3D%5Cfrac%7Bm_1RT_1%7D%7BMp_1%7D%5C%5C+m_1%3D%5Cfrac%7BVMp_1%7D%7BRT_1%7D%3D%5Cfrac%7B0%2C075%2A16%2A10%5E%7B-3%7D%2A2%2C4%2A10%5E6%7D%7B8%2C31%2A293%7D%3D1%2C2%5C%5C+%5Cfrac%7Bm_1%7D%7Bm_2%7D%3D5%2C8%5C%5C+m_2%3D%5Cfrac%7Bm_1%7D%7B5%2C8%7D%3D%5Cfrac%7B1%2C2%7D%7B5%2C8%7D%3D0%2C2%5C%5C+%5CDelta+m+%3Dm_1-m_2%3D1%2C2-0%2C2%3D1)
Ответ: 1 кг.