{"id":690,"date":"2026-10-02T08:36:06","date_gmt":"2026-10-02T08:36:06","guid":{"rendered":"https:\/\/giapsgk.com\/chuyen-de\/?p=690"},"modified":"2026-10-02T08:36:06","modified_gmt":"2026-10-02T08:36:06","slug":"giai-bai-tap-mo-hinh-dong-hoc-phan-tu-chat-khi-vat-li-12","status":"publish","type":"post","link":"https:\/\/giapsgk.com\/chuyen-de\/giai-bai-tap-mo-hinh-dong-hoc-phan-tu-chat-khi-vat-li-12\/","title":{"rendered":"Gi\u1ea3i b\u00e0i t\u1eadp m\u00f4 h\u00ecnh \u0111\u1ed9ng h\u1ecdc ph\u00e2n t\u1eed ch\u1ea5t kh\u00ed – V\u1eadt L\u00ed 12"},"content":{"rendered":"

Gi\u1ea3i b\u00e0i t\u1eadp m\u00f4 h\u00ecnh \u0111\u1ed9ng h\u1ecdc ph\u00e2n t\u1eed ch\u1ea5t kh\u00ed<\/strong> l\u00e0 ph\u1ea7n ki\u1ebfn th\u1ee9c quan tr\u1ecdng trong ch\u01b0\u01a1ng Kh\u00ed l\u00ed t\u01b0\u1edfng c\u1ee7a ch\u01b0\u01a1ng tr\u00ecnh V\u1eadt L\u00ed 12. \u0110\u1ec3 l\u00e0m t\u1ed1t d\u1ea1ng b\u00e0i n\u00e0y, b\u1ea1n c\u1ea7n n\u1eafm v\u1eefng m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa nhi\u1ec7t \u0111\u1ed9 v\u00e0 \u0111\u1ed9ng n\u0103ng t\u1ecbnh ti\u1ebfn trung b\u00ecnh c\u1ee7a ph\u00e2n t\u1eed ($E = \\frac{3}{2}kT$), c\u00f4ng th\u1ee9c t\u00ednh \u00e1p su\u1ea5t kh\u00ed theo m\u1eadt \u0111\u1ed9 \u0111\u1ed9ng n\u0103ng, c\u0169ng nh\u01b0 c\u00e1ch x\u00e1c \u0111\u1ecbnh t\u1ed1c \u0111\u1ed9 c\u0103n qu\u00e2n ph\u01b0\u01a1ng $\\sqrt{\\overline{v^2}} = \\sqrt{\\frac{3RT}{M}}$.<\/p>\n

B\u00e0i vi\u1ebft t\u1ed5ng h\u1ee3p c\u00e1c b\u00e0i t\u1eadp ti\u00eau bi\u1ec3u k\u00e8m \u0111\u00e1p \u00e1n v\u00e0 l\u1eddi gi\u1ea3i chi ti\u1ebft, g\u1ed3m: t\u00ednh \u0111\u1ed9ng n\u0103ng trung b\u00ecnh c\u1ee7a ph\u00e2n t\u1eed, t\u00ednh b\u00ecnh ph\u01b0\u01a1ng t\u1ed1c \u0111\u1ed9 v\u00e0 t\u1ed1c \u0111\u1ed9 c\u0103n qu\u00e2n ph\u01b0\u01a1ng, t\u00ednh \u00e1p su\u1ea5t t\u1eeb m\u1eadt \u0111\u1ed9 \u0111\u1ed9ng n\u0103ng, v\u00e0 x\u00e1c \u0111\u1ecbnh s\u1ed1 ph\u00e2n t\u1eed kh\u00ed trong b\u00ecnh. M\u1ed7i b\u00e0i \u0111\u1ec1u tr\u00ecnh b\u00e0y r\u00f5 c\u00f4ng th\u1ee9c, c\u00e1ch thay s\u1ed1 v\u00e0 \u0111\u01a1n v\u1ecb, gi\u00fap b\u1ea1n \u00f4n t\u1eadp hi\u1ec7u qu\u1ea3 v\u00e0 tr\u00e1nh c\u00e1c l\u1ed7i sai th\u01b0\u1eddng g\u1eb7p.<\/p>\n

C\u00e2u 1:<\/strong> \u0110\u1ed9ng n\u0103ng trung b\u00ecnh c\u1ee7a ph\u00e2n t\u1eed kh\u00ed Nitrogen \u1edf nhi\u1ec7t \u0111\u1ed9 $25^\\circ\\text{C}$ c\u00f3 gi\u00e1 tr\u1ecb l\u00e0 bao nhi\u00eau $\\times 10^{-21}$ J? (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn 2 ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n).<\/p>\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n

$E = \\frac{3}{2}kT = \\frac{3}{2}.1{,}38.10^{-23}.298 = 6{,}17.10^{-21}$<\/p>\n

C\u00e2u 2:<\/strong> \u1ede \u0111i\u1ec1u ki\u1ec7n ti\u00eau chu\u1ea9n (nhi\u1ec7t \u0111\u1ed9 l\u00e0 $0.00^\\circ\\text{C}$ v\u00e0 \u00e1p su\u1ea5t l\u00e0 $1.00$ atm), kh\u1ed1i l\u01b0\u1ee3ng mol c\u1ee7a kh\u00ed Oxygen l\u00e0 $32\\text{g\/mol}$. Gi\u00e1 tr\u1ecb trung b\u00ecnh c\u1ee7a b\u00ecnh ph\u01b0\u01a1ng t\u1ed1c \u0111\u1ed9 c\u00e1c ph\u00e2n t\u1eed kh\u00ed Oxygen l\u00e0 $X \\times 10^6\\ \\text{m}^2\/\\text{s}^2$. T\u00ecm X (vi\u1ebft k\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n 2 ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n). Cho $R = 8{,}314\\ J\/mol K$<\/p>\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n

$\\overline{v^2} = \\frac{3RT}{M} = \\frac{3kT}{m} = \\frac{3.8{,}314.273}{0{,}032} = 2{,}13.10^5 \\Rightarrow X \\approx 2{,}13$<\/p>\n

C\u00e2u 3:<\/strong> B\u00ecnh c\u00f3 dung t\u00edch $10.0$ l\u00edt ch\u1ee9a m\u1ed9t lo\u1ea1i kh\u00ed \u0111\u01a1n nguy\u00ean t\u1eed \u1edf nhi\u1ec7t \u0111\u1ed9 $400\\text{K}$ v\u00e0 \u00e1p su\u1ea5t $5 \\times 10^{-6}$ mmHg. S\u1ed1 ph\u00e2n t\u1eed kh\u00ed trong b\u00ecnh l\u00e0 $X \\times 10^{15}$. T\u00ecm X (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn 2 ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n).<\/p>\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n

$pV = nRT = \\frac{N}{N_A}RT \\Rightarrow N = \\frac{pVN_A}{RT} = \\frac{5.10^{-6}.133{,}32.10.10^{-3}.6{,}022.10^{23}}{8{,}314.400} = 1{,}21.10^{15}$<\/p>\n

$\\Rightarrow X = 1{,}21$<\/p>\n

C\u00e2u 4:<\/strong> X\u00e9t kh\u1ed1i kh\u00ed ch\u1ee9a trong m\u1ed9t b\u00ecnh k\u00edn, bi\u1ebft m\u1eadt \u0111\u1ed9 \u0111\u1ed9ng n\u0103ng ph\u00e2n t\u1eed (t\u1ed5ng \u0111\u1ed9ng n\u0103ng t\u1ecbnh ti\u1ebfn trung b\u00ecnh c\u1ee7a c\u00e1c ph\u00e2n t\u1eed kh\u00ed trong $1\\ \\text{m}^3$ th\u1ec3 t\u00edch kh\u00ed) c\u00f3 gi\u00e1 tr\u1ecb $5 \\times 10^5\\ \\text{J\/m}^3$. \u00c1p su\u1ea5t c\u1ee7a kh\u00ed trong b\u00ecnh b\u1eb1ng bao nhi\u00eau $\\times 10^5$ Pa? (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn 2 ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n).<\/p>\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n

\u0110\u1ed9ng n\u0103ng: $E = \\frac{3}{2}kT$<\/p>\n

M\u1eadt \u0111\u1ed9 \u0111\u1ed9ng n\u0103ng ph\u00e2n t\u1eed: $\\epsilon = \\frac{N}{V}.E = n.E$<\/p>\n

Theo \u0111\u1ecbnh lu\u1eadt kh\u00ed l\u00fd t\u01b0\u1edfng: $p = nkT = \\frac{2nE}{3} = \\frac{2}{3}\\epsilon = \\frac{2}{3}.5.10^5 = 3{,}33.10^3$<\/p>\n

C\u00e2u 5<\/strong>: T\u00ednh t\u1ed1c \u0111\u1ed9 c\u0103n qu\u00e2n ph\u01b0\u01a1ng (\u0111\u01a1n v\u1ecb m\/s) trong chuy\u1ec3n \u0111\u1ed9ng nhi\u1ec7t c\u1ee7a ph\u00e2n t\u1eed kh\u00ed Oxygen c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng mol l\u00e0 32 g\/mol \u1edf nhi\u1ec7t \u0111\u1ed9 400K. Coi c\u00e1c ph\u00e2n t\u1eed kh\u00ed l\u00e0 gi\u1ed1ng nhau. (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng \u0111\u01a1n v\u1ecb).<\/p>\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n

$\\overline{v^2} = \\frac{3RT}{M} \\Rightarrow \\bar{v} = \\sqrt{\\frac{3RT.N_A}{m}} = \\sqrt{\\frac{3RT}{M}} = \\sqrt{\\frac{3.8{,}314.400}{32.10^{-3}}} = 558$<\/p>\n

C\u00e2u 6<\/strong>: M\u1ed9t b\u00ecnh k\u00edn c\u00f3 th\u1ec3 t\u00edch $0.20\\ \\text{m}^3$ ch\u1ee9a kh\u00ed Neon \u1edf nhi\u1ec7t \u0111\u1ed9 $30^\\circ\\text{C}$ v\u00e0 \u00e1p su\u1ea5t $8.0 \\times 10^5$ Pa. Bi\u1ebft kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a ph\u00e2n t\u1eed kh\u00ed Neon l\u00e0 $m = 3.35 \\times 10^{-26}$ kg. M\u1ed9t trong c\u00e1c gi\u00e1 tr\u1ecb trung b\u00ecnh \u0111\u1eb7c tr\u01b0ng cho t\u1ed1c \u0111\u1ed9 c\u1ee7a c\u00e1c ph\u00e2n t\u1eed kh\u00ed th\u01b0\u1eddng d\u00f9ng l\u00e0 c\u0103n b\u1eadc hai c\u1ee7a trung b\u00ecnh b\u00ecnh ph\u01b0\u01a1ng t\u1ed1c \u0111\u1ed9 ph\u00e2n t\u1eed ($\\langle v^2 \\rangle$). Gi\u00e1 tr\u1ecb n\u00e0y c\u1ee7a c\u00e1c ph\u00e2n t\u1eed Neon trong b\u00ecnh l\u00e0 $X \\times 10^3$ m\/s. T\u00ecm X (vi\u1ebft k\u1ebft qu\u1ea3 ch\u1ec9 g\u1ed3m hai ch\u1eef s\u1ed1).<\/p>\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n

$\\overline{v^2} = \\frac{3RT}{M} \\Rightarrow \\bar{v} = \\sqrt{\\frac{3RTN_A}{m}} = 0{,}61.10^3$<\/p>\n

C\u00e2u 7<\/strong>: M\u1ed9t chi\u1ebfc xe t\u1ea3i v\u01b0\u1ee3t qua sa m\u1ea1c Atacama. Chuy\u1ebfn \u0111i b\u1eaft \u0111\u1ea7u v\u00e0o s\u00e1ng s\u1edbm khi nhi\u1ec7t \u0111\u1ed9 l\u00e0 $5.0^\\circ\\text{C}$. \u0110\u1ebfn gi\u1eefa tr\u01b0a, nhi\u1ec7t \u0111\u1ed9 t\u0103ng l\u00ean \u0111\u1ebfn $45^\\circ\\text{C}$. Coi kh\u00ed trong l\u1edbp xe c\u00f3 nhi\u1ec7t \u0111\u1ed9 nh\u01b0 ngo\u00e0i tr\u1eddi. \u0110\u1ed9 t\u0103ng \u0111\u1ed9ng n\u0103ng t\u1ecbnh ti\u1ebfn trung b\u00ecnh c\u1ee7a m\u1ed9t ph\u00e2n t\u1eed kh\u00ed do s\u1ef1 gia t\u0103ng nhi\u1ec7t \u0111\u1ed9 n\u00e0y c\u00f3 gi\u00e1 tr\u1ecb l\u00e0 bao nhi\u00eau $\\times 10^{-21}$ J? (K\u1ebft qu\u1ea3 l\u00e0m tr\u00f2n \u0111\u1ebfn 2 ch\u1eef s\u1ed1 th\u1eadp ph\u00e2n).<\/p>\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n

\u0110\u1ed9 t\u0103ng \u0111\u1ed9ng n\u0103ng t\u1ecbnh ti\u1ebfn trung b\u00ecnh:<\/p>\n

$\\Delta E = \\frac{3}{2}k(\\Delta T) = \\frac{3}{2}k(T_2 – T_1) = \\frac{3}{2}.1{,}38.10^{-23}.(318 – 278) = 8{,}28.10^{-22} \\approx 0{,}83.10^{-21}$<\/p>\n\n

\u2b07 T\u1ea3i \u0110\u1ec1<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Gi\u1ea3i b\u00e0i t\u1eadp m\u00f4 h\u00ecnh \u0111\u1ed9ng h\u1ecdc ph\u00e2n t\u1eed ch\u1ea5t kh\u00ed l\u00e0 ph\u1ea7n ki\u1ebfn th\u1ee9c quan tr\u1ecdng trong ch\u01b0\u01a1ng Kh\u00ed l\u00ed t\u01b0\u1edfng c\u1ee7a ch\u01b0\u01a1ng tr\u00ecnh V\u1eadt L\u00ed 12. \u0110\u1ec3 l\u00e0m t\u1ed1t d\u1ea1ng b\u00e0i n\u00e0y, b\u1ea1n c\u1ea7n n\u1eafm v\u1eefng m\u1ed1i li\u00ean h\u1ec7 gi\u1eefa nhi\u1ec7t \u0111\u1ed9 v\u00e0 \u0111\u1ed9ng n\u0103ng t\u1ecbnh ti\u1ebfn trung b\u00ecnh c\u1ee7a ph\u00e2n t\u1eed ($E … \u0110\u1ecdc ti\u1ebfp<\/a><\/p>\n","protected":false},"author":1,"featured_media":693,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":["post-690","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-bai-tap"],"_links":{"self":[{"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/posts\/690","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/comments?post=690"}],"version-history":[{"count":1,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/posts\/690\/revisions"}],"predecessor-version":[{"id":695,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/posts\/690\/revisions\/695"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/media\/693"}],"wp:attachment":[{"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/media?parent=690"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/categories?post=690"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/giapsgk.com\/chuyen-de\/wp-json\/wp\/v2\/tags?post=690"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}