A) \[1.6\times {{10}^{5}}N{{m}^{-2}}\]
B) \[1.8\times {{10}^{6}}N{{m}^{-2}}\]
C) \[3.6\times {{10}^{6}}N{{m}^{-2}}\]
D) \[8.1\times {{10}^{3}}N{{m}^{-2}}\]
Correct Answer: C
Solution :
| We know that \[PV=nRT\] |
| \[\therefore \,\,n=\frac{PV}{RT}=\frac{1.6\times {{10}^{6}}\times 0.0083}{8.3\times 300}=\frac{16}{3}=5.33\,moles\] \[{{C}_{P}}=\frac{5R}{2}\Rightarrow {{C}_{V}}=\frac{3R}{2}\] |
| When \[2.49\times {{10}^{4}}J\] of heat energy is supplied at constant volume then we can use the following relationship to find change in temperature. |
| \[Q=n{{C}_{V}}\Delta T\] |
| \[\therefore \,\,\,\Delta T=\frac{Q}{n{{C}_{V}}}=\frac{2.49\times {{10}^{4}}}{5.33\times \frac{3}{2}\times 8.3}=375K\] |
| Therefore, the final temperature |
| \[=300+375=675K\] |
| Applying Gay Lussac's Law, to find pressure. |
| \[\frac{{{P}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}}{{{T}_{2}}}\] |
| \[\Rightarrow \,\,\,{{P}_{2}}=\frac{{{P}_{1}}{{T}_{2}}}{{{T}_{1}}}=\frac{1.6\times {{10}^{6}}\times 675}{300}=3.6\times {{10}^{6}}N{{m}^{-2}}\] |
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