The volume of 0.354g of helium at 273°C and 114cmhg of mercury pressure is 2667cm\(^3\). Calculate the volume
2000.25 cm\(^3\)
200.025 cm\(^3\)
4003.27 cm\(^3\)
2010.25 cm\(^3\)
Correct answer is A
m = 0.354g, T\(_1\) = 273°C = 273 + 273 = 576K
P\(_1\) = 114cmHg, V\(_1\) = 2667cm\(^3\) at STP
T\(^2\) = 273K, P\(_2\) = 76cmHg, V\(_2\) = ?
| P\(^1\)V\(^1\)
T\(^1\) |
= | P\(^2\)V\(^2\)
T\(^1\) |
| V\(_2\) | = | 114 × 2667 × 273
76 × 576 |
= | 2000.25cm\(^3\) |