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{"id":44180,"date":"2019-10-13T14:16:34","date_gmt":"2019-10-13T07:16:34","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=44180"},"modified":"2019-10-13T14:16:34","modified_gmt":"2019-10-13T07:16:34","slug":"giai-vat-li-10-bai-31-phuong-trinh-trang-thai-cua-khi-li-tuong","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/giai-vat-li-10-bai-31-phuong-trinh-trang-thai-cua-khi-li-tuong\/","title":{"rendered":"Gi\u1ea3i V\u1eadt L\u00ed 10 B\u00e0i 31 : Ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00ed t\u01b0\u1edfng"},"content":{"rendered":"\n

B\u00e0i 31 : Ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00ed t\u01b0\u1edfng<\/h2>\n\n\n\n

C1.<\/strong> ( trang 160 sgk V\u1eadt L\u00fd 10): – L\u01b0\u1ee3ng \nkh\u00ed \u0111\u01b0\u1ee3c chuy\u1ec3n t\u1eeb tr\u1ea1ng th\u00e1i 1 sang tr\u1ea1ng th\u00e1i 1′ b\u1eb1ng qu\u00e1 tr\u00ecnh n\u00e0o? \nH\u00e3y vi\u1ebft bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa p1<\/sub>, V1<\/sub> v\u00e0 p’, V2<\/sub>.<\/p>\n\n\n\n

– L\u01b0\u1ee3ng kh\u00ed \u0111\u01b0\u1ee3c chuy\u1ec3n t\u1eeb tr\u1ea1ng th\u00e1i 1′ sang tr\u1ea1ng th\u00e1i 2 b\u1eb1ng qu\u00e1 tr\u00ecnh n\u00e0o? H\u00e3y vi\u1ebft bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa p’, T1<\/sub> v\u00e0 p2<\/sub>, T2<\/sub>.<\/ins><\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi:<\/strong><\/p>\n\n\n\n

+ Tr\u1ea1ng th\u00e1i (1) sang tr\u1ea1ng th\u00e1i (1\u2019) l\u00e0 qu\u00e1 tr\u00ecnh \u0111\u1eb3ng nhi\u1ec7t v\u00ec nhi\u1ec7t \u0111\u1ed9 T1<\/sub> \u0111\u01b0\u1ee3c gi\u1eef nguy\u00ean. Bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7: p1<\/sub>.V1<\/sub> = p\u2019.V2<\/sub> (I).<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

+ Tr\u1ea1ng th\u00e1i (1\u2019) sang tr\u1ea1ng th\u00e1i (2) l\u00e0 qu\u00e1 tr\u00ecnh \u0111\u1eb3ng t\u00edch v\u00ec th\u1ec3 t\u00edch V2<\/sub> \u0111\u01b0\u1ee3c gi\u1eef nguy\u00ean. Bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7: \n (II)<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

+ T\u1eeb (I) suy ra: \n th\u1ebf v\u00e0o (II), ta \u0111\u01b0\u1ee3c: \n <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Hay: . \u0110\u00e2y l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00ed t\u01b0\u1edfng.<\/p>\n\n\n\n

B\u00e0i 4 (trang 165 SGK V\u1eadt L\u00fd 10) :<\/strong> H\u00e3y gh\u00e9p c\u00e1c qu\u00e1 tr\u00ecnh ghi b\u00ean tr\u00e1i v\u1edbi c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u1ee9ng ghi b\u00ean ph\u1ea3i.<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n
1 – c<\/td>2 – a<\/td>3 – b<\/td>4 – d<\/td><\/tr><\/tbody><\/table>\n\n\n\n

Ch\u00fa \u00fd: C\u00f4ng th\u1ee9c (d) \u00e1p d\u1ee5ng cho qu\u00e1 tr\u00ecnh bi\u1ebfn \u0111\u1ed5i b\u1ea5t k\u00ec tr\u1ea1ng th\u00e1i ch\u1ea5t kh\u00ed l\u00fd t\u01b0\u1edfng nh\u01b0ng \u0111i\u1ec1u ki\u1ec7n l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng ch\u1ea5t kh\u00ed kh\u00f4ng \u0111\u1ed5i trong su\u1ed1t qu\u00e1 tr\u00ecnh x\u1ea3y ra bi\u1ebfn \u0111\u1ed5i tr\u1ea1ng th\u00e1i.<\/p>\n\n\n\n

B\u00e0i 5 (trang 166 SGK V\u1eadt L\u00fd 10) :<\/strong> Trong h\u1ec7 t\u1ecda \u0111\u1ed9 (V, T), \u0111\u01b0\u1eddng bi\u1ec3u di\u1ec5n n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u01b0\u1eddng \u0111\u1eb3ng \u00e1p?<\/ins><\/p>\n\n\n\n

A. \u0110\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c ho\u00e0nh<\/p>\n\n\n\n

B. \u0110\u01b0\u1eddng th\u1eb3ng song song v\u1edbi tr\u1ee5c tung<\/p>\n\n\n\n

C. \u0110\u01b0\u1eddng hypebol<\/p>\n\n\n\n

D. \u0110\u01b0\u1eddng th\u1eb3ng k\u00e9o d\u00e0i \u0111i qua g\u1ed1c t\u1ecda \u0111\u1ed9.<\/p>\n\n\n\n

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

Ch\u1ecdn D.<\/p>\n\n\n\n

B\u00e0i 6 (trang 166 SGK V\u1eadt L\u00fd 10) :<\/strong> M\u1ed1i \nli\u00ean h\u1ec7 gi\u1eefa \u00e1p su\u1ea5t th\u1ec3 t\u00edch, nhi\u1ec7t \u0111\u1ed9 c\u1ee7a m\u1ed9t l\u01b0\u1ee3ng kh\u00ed trong qu\u00e1 \ntr\u00ecnh n\u00e0o sau \u0111\u00e2y kh\u00f4ng \u0111\u01b0\u1ee3c x\u00e1c \u0111\u1ecbnh b\u1eb1ng ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a \nkh\u00ed l\u00ed t\u01b0\u1edfng?<\/ins><\/p>\n\n\n\n

A. Nung n\u00f3ng m\u1ed9t l\u01b0\u1ee3ng kh\u00ed trong m\u1ed9t b\u00ecnh \u0111\u1eady k\u00edn<\/p>\n\n\n\n

B. Nung n\u00f3ng m\u1ed9t l\u01b0\u1ee3ng kh\u00ed trong m\u1ed9t b\u00ecnh kh\u00f4ng \u0111\u1eady k\u00edn<\/p>\n\n\n\n

C. Nung n\u00f3ng m\u1ed9t l\u01b0\u1ee3ng kh\u00ed trong m\u1ed9t xilanh k\u00edn c\u00f3 pit-t\u00f4ng l\u00e0m kh\u00ed n\u00f3ng l\u00ean, n\u1edf ra, \u0111\u1ea9y pit-t\u00f4ng di chuy\u1ec3n<\/p>\n\n\n\n

D. D\u00f9ng tay b\u00f3p l\u00f5m qu\u1ea3 b\u00f3ng b\u00e0n.<\/p>\n\n\n\n

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

Chon B. V\u00ec khi nung n\u00f3ng m\u00e0 b\u00ecnh kh\u00f4ng \u0111\u1eady k\u00edn, m\u1ed9t l\u01b0\u1ee3ng kh\u00ed s\u1ebd tho\u00e1t ra ngo\u00e0i, ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i s\u1ebd kh\u00f4ng \u0111\u01b0\u1ee3c nghi\u1ec7m \u0111\u00fang.<\/p>\n\n\n\n

B\u00e0i 7 (trang 166 SGK V\u1eadt L\u00fd 10) :<\/strong> Trong ph\u00f2ng th\u00ed nghi\u1ec7m, ng\u01b0\u1eddi ta \u0111i\u1ec1u ch\u1ebf \u0111\u01b0\u1ee3c 40 cm3<\/sup> kh\u00ed hidro \u1edf \u00e1p su\u1ea5t 750 mmHg v\u00e0 nhi\u1ec7t \u0111\u1ed9 27o<\/sup> C. T\u00ednh th\u1ec3 t\u00edch c\u1ee7a l\u01b0\u1ee3ng kh\u00ed tr\u00ean \u1edf \u0111i\u1ec1u ki\u1ec7n chu\u1ea9n (\u00e1p su\u1ea5t 760 mmHg v\u00e0 nhi\u1ec7t \u0111\u1ed9 0o<\/sup> C)<\/p>\n\n\n\n

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

Tr\u1ea1ng th\u00e1i 1:<\/p>\n\n\n\n

       P1<\/sub> = 750 mmHg<\/p>\n\n\n\n

       T1<\/sub> = 27 + 273 = 300 K<\/p>\n\n\n\n

       V1<\/sub> = 40 cm3<\/sup><\/p>\n\n\n\n

Tr\u1ea1ng th\u00e1i 2:<\/p>\n\n\n\n

       Po<\/sub> = 760 mmHg<\/p>\n\n\n\n

       To<\/sub> = 0 + 273 = 273 K<\/p>\n\n\n\n

       Vo<\/sub> = ?<\/p>\n\n\n\n

\u00c1p d\u1ee5ng ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00ed t\u01b0\u1edfng:<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

B\u00e0i 8 (trang 166 SGK V\u1eadt L\u00fd 10) :<\/strong> T\u00ednh \nkh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a kh\u00f4ng kh\u00ed \u1edf \u0111\u1ec9nh n\u00fai Ph\u0103ng-xi-p\u0103ng cao 3 140 m. \nBi\u1ebft r\u1eb1ng m\u1ed7i khi l\u00ean cao th\u00eam 10m th\u00ec \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n gi\u1ea3m 1 mmHg v\u00e0 \nnhi\u1ec7t \u0111\u1ed9 tr\u00ean \u0111\u1ec9nh n\u00fai l\u00e0 2o<\/sup> C. Kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a kh\u00f4ng kh\u00ed \u1edf \u0111i\u1ec1u ki\u1ec7n ti\u00eau chu\u1ea9n (\u00e1p su\u1ea5t 760 mmHg v\u00e0 nhi\u1ec7t \u0111\u1ed9 0o<\/sup> C) l\u00e0 1,29 kg\/m3<\/sup>.<\/ins><\/p>\n\n\n\n

– Tr\u1ea1ng th\u00e1i 1 (chu\u1ea9n) <\/p>\n\n\n\n

       Po<\/sub> = 760 mmHg <\/p>\n\n\n\n

       To<\/sub> = 0 + 273 = 273 K <\/p>\n\n\n\n

       Vo<\/sub> = ? <\/p>\n\n\n\n

– Tr\u1ea1ng th\u00e1i 2 (\u1edf \u0111\u1ec9nh n\u00fai)<\/p>\n\n\n\n

       P = (760 \u2013 314) mmHg<\/p>\n\n\n\n

       T = 275 K<\/p>\n\n\n\n

       V = ?<\/p>\n\n\n\n

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

\"Gi\u1ea3i<\/figure>\n\n\n\n

Khi l\u00ean cao th\u00eam 10m th\u00ec \u00e1p su\u1ea5t kh\u00ed quy\u1ec3n gi\u1ea3m 1 mmHg. Do \u0111\u00f3 l\u00ean cao 3140m, \u00e1p su\u1ea5t kh\u00f4ng kh\u00ed gi\u1ea3m: \n<\/p>\n\n\n\n

\u2192 \u00c1p su\u1ea5t kh\u00f4ng kh\u00ed \u1edf tr\u00ean \u0111\u1ec9nh n\u00fai Ph\u0103ng-xi-p\u0103ng: p1<\/sub> = 760 \u2013 314 = 446 mmHg<\/p>\n\n\n\n

Kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a kh\u00f4ng kh\u00ed: <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

\u00c1p d\u1ee5ng ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i ta \u0111\u01b0\u1ee3c:<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Kh\u1ed1i l\u01b0\u1ee3ng ri\u00eang c\u1ee7a kh\u00f4ng kh\u00ed \u1edf \u0111\u1ec9nh n\u00fai Ph\u0103ng-xi-p\u0103ng cao 3 140 m:<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 31 : Ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00ed t\u01b0\u1edfng C1. ( trang 160 sgk V\u1eadt L\u00fd 10): – L\u01b0\u1ee3ng kh\u00ed \u0111\u01b0\u1ee3c chuy\u1ec3n t\u1eeb tr\u1ea1ng th\u00e1i 1 sang tr\u1ea1ng th\u00e1i 1′ b\u1eb1ng qu\u00e1 tr\u00ecnh n\u00e0o? H\u00e3y vi\u1ebft bi\u1ec3u th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa p1, V1 v\u00e0 p’, V2. – L\u01b0\u1ee3ng kh\u00ed \u0111\u01b0\u1ee3c chuy\u1ec3n t\u1eeb tr\u1ea1ng […]<\/p>\n","protected":false},"author":12,"featured_media":44181,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1640],"tags":[],"yoast_head":"\nGi\u1ea3i V\u1eadt L\u00ed 10 B\u00e0i 31 : Ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00ed t\u01b0\u1edfng<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/giai-vat-li-10-bai-31-phuong-trinh-trang-thai-cua-khi-li-tuong\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Gi\u1ea3i V\u1eadt L\u00ed 10 B\u00e0i 31 : Ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00ed t\u01b0\u1edfng\" \/>\n<meta property=\"og:description\" content=\"B\u00e0i 31 : Ph\u01b0\u01a1ng tr\u00ecnh tr\u1ea1ng th\u00e1i c\u1ee7a kh\u00ed l\u00ed t\u01b0\u1edfng C1. 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