The volume of 0.354g of helium at 273°C and 114cm of mercury pressure is 2667cm\(^3\). Calculate the volume
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
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\(^1\)
Â
V\(_2\)
=
114 × 2667 × 273
76 × 576
=
2000.25cm\(^3\)
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\(^1\)
Â
V\(_2\)
=
114 × 2667 × 273
76 × 576
=
2000.25cm\(^3\)