Đề thi thử tốt nghiệp THPT môn Vật lí THPT Phú Nhuận - HCM có đáp án

$mc(t{}_{1}-t{}_{cb})=m{}_{2}c(t{}_{cb}-t{}_{2})\Rightarrow m(80-t{}_{cb})=5(t{}_{cb}-20)\Rightarrow m=\frac{5(t{}_{cb}-20)}{80-t{}_{cb}}$

$mc(θ{}_{cb}-t{}_{cb})=(m{}_{1}-m)c(t{}_{1}-θ{}_{cb})\Rightarrow \frac{5(t{}_{cb}-20)}{80-t{}_{cb}}\cdot (77{,}92-t{}_{cb})=[3-\frac{5(t{}_{cb}-20)}{80-t{}_{cb}}](80-77{,}92)$

$\Rightarrow t{}_{cb}\approx 21{,}2{}^{^{\circ}}\text{C}.$Chọn A

a) Đúng

b) Sai. $A<0$

c) Đúng

d) Sai. $Q>0$

$Q{}_{AB}=mcΔt=2.4180.100=836000\text{J}=836\text{k}\text{J}$

a) Đúng; b) Đúng; c) Đúng; d) Sai

$W{}_{d}=\frac{3}{2}kT\Rightarrow$ a) Đúng

$p=\frac{1}{3}μm\overline{v{}^{2}}\Rightarrow p\sim \overline{v{}^{2}}\Rightarrow$ b) Sai

$μ=\frac{N}{V}=\frac{9{,}68\cdot 10{}^{20}}{2\cdot 10{}^{-6}}=4{,}84\cdot 10{}^{26}\text{m}{}^{-3}$

$W{}_{d}=\frac{3}{2}kT=\frac{3p}{2μ}=\frac{3\cdot 4\cdot 10{}^{6}}{2\cdot 4{,}84\cdot 10{}^{26}}\approx 12{,}4\cdot 10{}^{-21}\text{J}\Rightarrow$ c) Đúng

$\Rightarrow \frac{3}{2}\cdot 1{,}38\cdot 10{}^{-23}\cdot T=12{,}4\cdot 10{}^{-21}\Rightarrow T\approx 599K\Rightarrow$ d) Đúng

Quá trình (3)-(4) không đi qua gốc tọa độ nên không phải đẳng tích $\Rightarrow$ a) Sai

Quá trình (2)-(3) là đẳng áp $\frac{V}{T}=$ const $\Rightarrow T↑$ thì $V↑\Rightarrow$ b) Đúng

Quá trình (1)-(2) là đẳng tích $\frac{p{}_{1}}{T{}_{1}}=\frac{p{}_{2}}{T{}_{2}}\Rightarrow \frac{1}{290}=\frac{1{,}8}{T{}_{2}}\Rightarrow T{}_{2}=522\text{K}\Rightarrow$ c) Đúng

Hình bình hành $\Rightarrow T{}_{3}-T{}_{2}=T{}_{4}-T{}_{1}\Rightarrow T{}_{3}-522=870-290\Rightarrow T{}_{3}=1102\text{K}$

$\frac{p{}_{1}V{}_{1}}{T{}_{1}}=\frac{p{}_{3}V{}_{3}}{T{}_{3}}\Rightarrow \frac{1.400}{290}=\frac{1{,}8.V{}_{3}}{1102}\Rightarrow V{}_{3}\approx 844\text{c}\text{m}{}^{3}\Rightarrow$d) Sai

$T=t+273=37+273=310(K)$

Đáp án: 310

$Q{}_{B}-Q{}_{A}=mcΔt\Rightarrow (375-165)\cdot 10{}^{3}=m\cdot 4200\cdot 100\Rightarrow m=0{,}5\text{k}\text{g}$

Đáp án: 0,5

$P=UI=1{,}65.2{,}5=4{,}125W$

$Q=Pt=4{,}125\cdot (8\cdot 60+48)=2178\text{J}$

$Q=mcΔt\Rightarrow 2178=0{,}15\cdot c\cdot (65{,}5-62)\Rightarrow c\approx 4149\text{J}/\text{k}\text{g}\text{K}$

Đáp án: 4149

Đẳng áp $\frac{V}{T}=$ const $\Rightarrow \frac{1}{31+273}=\frac{0{,}5}{T}\Rightarrow T=152\text{K}=-121{}^{^{\circ}}\text{C}$

Đáp án: - 121

$U=\frac{i}{2}nRT=\frac{i}{2}pV\overset{U=U{}_{1}+U{}_{2}}{\rightarrow }pV=p{}_{1}V{}_{1}+p{}_{2}V{}_{2}\Rightarrow p.4=430.1+940.3\Rightarrow p=812{,}5\text{m}\text{m}\text{H}\text{g}$

$n=\frac{pV}{RT}\overset{n=n{}_{1}+n{}_{2}}{\rightarrow }\frac{pV}{RT}=\frac{p{}_{1}V{}_{1}}{RT{}_{1}}+\frac{p{}_{2}V{}_{2}}{RT{}_{2}}\Rightarrow \frac{812{,}5.4}{T}=\frac{430.1}{443}+\frac{940.3}{483}\Rightarrow T\approx 477K\approx 204{}^{o}\text{C}$

Đáp án: 204

$\{\begin{matrix}pV=(p+2.10{}^{5})(V-3) \\ pV=(p+5.10{}^{5})(V-5)\end{matrix}$

$\Rightarrow p=4.10{}^{5}\text{P}\text{a}=400\text{k}\text{P}\text{a}$

Đáp án: 400