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

$\frac{p}{T}=$ const. Chọn C

Chọn C

Chọn D

Bình kín thì thể tích không đổi $\Rightarrow A=0$. Chọn A

$\frac{pV}{T}=C\Rightarrow V=\frac{C}{p}\cdot T\Rightarrow$ hệ số góc $\frac{C}{p{}_1}>\frac{C}{p{}_2}\Rightarrow p{}_1<p{}_2.$Chọn C

$mc\left(t{}_1-t{}_{cb}\right)=m{}_{2}c\left(t{}_{cb}-t{}_2\right)\Rightarrow m\left(80-t{}_{cb}\right)=5\left(t{}_{cb}-20\right)\Rightarrow m=\frac{5\left(t{}_{cb}-20\right)}{80-t{}_{cb}}$

$mc\left(\theta {}_{cb}-t{}_{cb}\right)=\left(m{}_1-m\right)c\left(t{}_1-\theta {}_{cb}\right)\Rightarrow \frac{5\left(t{}_{cb}-20\right)}{80-t{}_{cb}}\cdot \left(77,92-t{}_{cb}\right)=\left[3-\frac{5\left(t{}_{cb}-20\right)}{80-t{}_{cb}}\right](80-77,92)$

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

$v{}_c=\sqrt{\frac{3kT}{m}}\Rightarrow \frac{v{}_{cA}}{v{}_{cB}}=\sqrt{\frac{m{}_B}{m{}_A}}=2\Rightarrow \frac{m{}_B}{m{}_A}=4.$Chọn A

Đẳng áp $\frac{V}{T}=$ const $\Rightarrow \frac{10}{27+273}=\frac{10+\Delta V}{27+120+273}\Rightarrow \Delta V=4l$. Chọn B

$\frac{p}{DT}=\text{c}\text{o}\text{n}\text{s}\text{t}\Rightarrow \frac{760}{1,29\cdot 273}=\frac{760-\frac{3143}{10}}{D\cdot (2+273)}\Rightarrow D\approx 0,75 \text{k}\text{g}/\text{m}{}^3.$Chọn A

a) Đúng

b) Sai. $A<0$

c) Đúng

d) Sai. $Q>0$

$Q{}_{AB}=mc\Delta t=2.4180.100=836000 \text{J}=836 \text{k}\text{J}$

a) Đúng; b) Đúng; c) Đúng; d) Sai

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

$p=\frac{1}{3}\mu m\overline{v{}^2}\Rightarrow p~\overline{v{}^2}\Rightarrow$ b) Sai

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

$W{}_d=\frac{3}{2}kT=\frac{3p}{2\mu }=\frac{3\cdot 4\cdot 10{}^6}{2\cdot 4,84\cdot 10{}^{26}}\approx 12,4\cdot 10{}^{-21} \text{J}\Rightarrow$ 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 $\Rightarrow T\uparrow$ thì $V\uparrow \Rightarrow$ 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

$Q{}_B-Q{}_A=mc\Delta t\Rightarrow (375-165)\cdot 10{}^3=m\cdot 4200\cdot 100\Rightarrow m=0,5 \text{k}\text{g}$

Đáp án: 0,5

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

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

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

Đáp án: 4149

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

Đáp án: - 121

$U=\frac{i}{2}nRT=\frac{i}{2}pV\stackrel{U=U{}_1+U{}_2}{\to }pV=p{}_{1}V{}_1+p{}_{2}V{}_2\Rightarrow p.4=430.1+940.3\Rightarrow p=812,5\text{m}\text{m}\text{H}\text{g}$

$n=\frac{pV}{RT}\stackrel{n=n{}_1+n{}_2}{\to }\frac{pV}{RT}=\frac{p{}_{1}V{}_1}{RT{}_1}+\frac{p{}_{2}V{}_2}{RT{}_2}\Rightarrow \frac{812,5.4}{T}=\frac{430.1}{443}+\frac{940.3}{483}\Rightarrow T\approx 477K\approx 204{}^o\text{C}$

Đáp án: 204

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

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

Đáp án: 400